Though I have a gist of time and length change due to the point of view by an observer, I do not know the equations on how to work out the actual details.

Now to help with this, I have hypothetical scenario in which two space craft head in opposite directions (solar system orbit around the center of the galaxy).  

The hypothetical mission is to place two telescopes in orbit around the center of the galaxy, one leading the sun/solar system, and the other trailing it.  Both spacecraft are manned. A propulsion device has been created that will provide constant acceleration at 10m/s² for four years.  The plan is for each to provide thrust for one year, coast for 6 months, then provide thrust in the opposite direction for a year (stopping relative to the sun), taking a week to position the telescope, then return to Earth the same way.

Now the way I understand it (I could be quite wrong), the crews will feel the slightly more than one 'g' and the distance of actual travel can be calculated using this acceleration, coast, and negative acceleration.  Also the crews would feel that they were gone from earth five years and a week (plus the time to get out in orbit, and back).  However, the perception of those on earth is where I am really fuzzy.  Also, how does our speed of travel around the center of the galaxy affect calculations?

I don’t quite understand what you are trying to say, starting off talking about four years at one g.
But, with regards to some of your specifics,
You’re talking about heading from here towards Sagittarius A in the center of the Galaxy, about 26K ly away.  
If you had an engine that could provide a constant 1g thrust for a year,
That would take about 20 years to do for guys on the ship, about 30K years for people on earth, some 10 ton spacecraft would weigh like 8B tons, and you would be going like 0.99999999999 the speed of light.

So let’s cut back to what would happen with a year of the thrust.
You would not make it to our closest other star.
You would have headed off towards Alpha Centauri-
Which is a little over 4ly away.
That would take about 3.5 years to reach.  About six years would have gone by on earth.  You ship would weigh like 400 tons.  You would be moving about 0.95 the speed of light.  

Now, in a Newtonian universe, I guess you could be reaching Star Trek speeds like warp 2, warp 4, etc. assuming that means twice or four times the speed of light.  But for quick Einstein you can’t exceed the speed of light punched in to the HP15C real quick the numbers above should be fairly close.

This also requires universal inertia.
Like, the earth is rotating at about 1000 mph.  Maybe at Idaho at 45N or so with the cosine about 750mph.
So, if you set down your lap top, stand up, then jump in the air, the wall of your house does not smack you at 750 mph.

And, as the earth rotates around the sun at 67,000 mph
And when you get in your space ship, and take off - the earth does not disappear from you at 67K mph.

And when you get out of our solar system it does not disappear at the 50K mph it rotates the center of the galaxy.

(And let’s not get into galaxy spin, rotation about the center of the universes, cluster movement with regards to other clusters, etc.)

It’s kind of a party pooper.
Right now technologically we can produce a few G for a few minutes.
Not remotely anything like a constant 1g.
And…
Even at a small fraction of light speed,  even 1% is like 6.7 million mph.
What happens when your shit hits a gram of iron or ice at that speed?
Or even less?
A 31 grain/2 gram 22 long rifle at 1200mph is probably something most of you can relate to.
It will Jack up all kinds of metals and plastics.
So, what would something hit at 5000 to 6000 times faster do?

Let's see if I can make my questions a bit more clear.

When, from Earth's perspective, as the space craft moves faster, the folks on board become more massive.  However, inversely proportional to that mass gain, the acceleration slows (time), which keeps the occupant's weight constant.  From the perspective of those on board, the acceleration is constant until the engine is shut off.  Even form the engine's perspective, it is still providing the same thrust, but due to time dilation its thrust from Earth's perspective is reducing. When the engine is shut off, they are coasting weightlessly for six months.  Now, to bring their speed to the same as the sun around the galaxy center, they have to again accelerate in the opposite direction.  Again from their perspective, they are at a constant weight and mass, though from Earth, they are reducing mass and slowing at a greater rate.

First question: what are the equations that express the proportional time and mass with respect to speed from the "stationary" observer?  Second question: How does the time between the two space craft differ with respect to Earth  Third, is the distance where the telescopes wind up the same as when calculated by using Newtonian laws of motion?

Obviously, the Earth is not stationary, nor is the sun.  So, the part about putting one telescope leading the sun's orbit, and the other trailing, is to compare how the solar system's speed enters the problem.  That is my next question.  What are the relationships to the two space craft relative to Earth with respect to the occupant's age upon returning.


ETA:  If my interpretation is all wrong or partly wrong, please correct me.

Time dilation, mass increase, etc., all operate under the same formula:  1/((1-v²/c²)^1/2)

In this case, v is the velocity with respect to the "stationary observer," and c is the speed of light.  At v=0, the above formula returns 1; mass is "rest mass," and time passes normally.  As v approaches c, the result rises to infinity - mass increases toward infinity, and the passage of time slows toward zero (1/infinity).

The reason you can never reach the speed of light is because as you try to accelerate toward the speed of light, more and more of the added energy simply increases the mass (aka "relativistic mass") rather than speeding it up.  Just as you can never reach infinity, you can also never accelerate to the speed of light.  Photons travel at the speed of light because they have zero rest mass - they have no mass and therefore no energy at any velocity other than the speed of light.

The heart of Special Relativity is the Lorentz transformation. Study this first, then everything else will become clearer.

Again,
Your questions are confusing.

the equations are some pretty basic shit that can make the concepts clear in a non Calc based HS physics class.
They literally make little graphs and pictures for that audience with the equations printed out in them.
You don’t need to go on some big quest on a forum.

Same with the time thing.  
Literal one page comic book looking age/time comparisons.

What you are talking about calculating is not “Newtonian.”  

For a basic simplification-

Newtonian/classical physics is how we drop artillery on somebody, go to the moon, get a lander on mars, etc.
Under Newtonian physics, I could hop in a space ship, punch it to 1G, and snap a picture of the closest star four light years away as I was passing it, under 3 years from now, going 3X the speed of light.
But it does not really work like that.

When shit gets really, really fast-  relativity comes into play.
I would hop in my spaceship, punch it to one G, and pass that star around 2 years from now, going 0.98 the speed of light,
While more like five years have gone by for you here on earth.

And this is just basic getting there as you drive by.  not adding coasting, slowing, return etc.

When shit gets really, really small it does not work that way either.  There is no calculating exactly where anything is, it’s a possibility of where it might be prediction.  That’s quantum physics.

When shit is both very, very small and very, very fast-
That’s quantum field theory.

Now, when shit is really super, very, extra big-like a galaxy-
The way the various stars and stuff behave and rotate and the way galaxies interact with each other is not Newtonian.
So either, really big shit not at all the way it should be, or is maybe kind of sort of Newtonian if dark matter exists.

Right know, your questions are sort of along the lines of-
“How much is a tax stamp on a machine gun?  How do you file the firing pin to make a machine gun? When my machine gun magazine empties and the slide is clacking back and forth on empty, is it at the same rounds per minute as when I was firing?  Why do my bullets rise when I shoot them?”

So it’s kind of hard to sort out.

Go read The Fabric of the Cosmos by Brian Green. He's an astrophysicist and professor. It's a really good book. Minimal math and he's able to distill a lot of this down pretty well. Most of the book is about the Big Bang, but there's a lot of relativity discussion too, as they have a lot of overlap.

I don’t well remember formulas, so I tend to remember this.

If you have a laser in the spaceship that emits photons perpendicular to the direction of travel of the spaceship (as seen by an observer onboard), how long does it take those photons to hit the opposite wall? If the distance to the wall is d, it is d/c .

To an outside observer things are different. The photons are still traveling at c, but they are not traveling perpendicular to the ships travel. Instead they have a velocity component v1 equal to the ships speed in the direction of the ships travel and another velocity component v2 perpendicular to the ships velocity. The total speed of the photon is of course still c (in a direction that will take it from the laser emitter to where the wall directly opposite the laser emitter will be by the time the photon gets there, making the hypotenuse of a right triangle), thus by Pythagoras theorem:

c^2 = v1^2 + v2^2
v2^2 = c^2 - v1^2
v2 = (c^2 - v1^2)^(1/2)

To an outside observer it takes d/v2 time for the photon emitted by the laser to hit the wall vs d/c for someone onboard the ship (since some of it’s speed is “wasted” keeping up with the ship so it still lands on the same spot on the opposite wall as seen by someone onboard the ship). So t’ = c / v2. Replace v2 with the above and rename v1 to v:

t’ = c / (c^2 - v^2)^(1/2)

This is very similar to the Lorentz formula mentioned above.

So if you are traveling as 0.5c your time dilation is

t’ = c / (c^2 - (c/2)^2)^(1/2)
t’ = c / (3c^2 / 4)^(1/2)
t’ = c / 0.866c
t’ = ~1.1547

So 1.1547 times as much time passes to an outside observer vs someone onboard the ship.