Trying to understand tensions Here's a picture of a problem I have:

I apologize for my drawing skills.
There's no friction in the problem, the strings and pulley are massless. You can assume box A is being held motionless, so there's no acceleration in the system. The only force present is gravity on mass m.
As you can see, I've determined the magnitude of the tension in string 2 to be equal to mg at mass m.
Now, from my understanding, force T at mass m is equal to force T at box B. Is this correct?
This is where I'm confused - do I take the mass of box B into account for the tension in string 1? How does the tension in string 2 relate to the tension in string 1?
 A: Box A is held motionless, and box B and mass $m$ are also motionless. 
None of the masses are accelerating $(a=0)$ so applying $F=ma$ to each the total force $F$ on each is zero. Therefore for mass $m$ and box B we can write :
$mg-T_2=0$
$T_2-T_1=0$.
So $T_1=T_2=mg$.
The mass of box B does not come into the problem because it is not accelerating, and because its weight does not act horizontally, in the direction of the strings.
A: Add mass A and B together and get the acc'n from F=ma.
Once you have the acc'n (they both have the same acc'n) you can get the individual force for mass A. 
A: Consider Tension like an opposite force acting against the direction in which a force is applied. 
Are you aware of the concept of a Normal force?
The easiest way to understand a force problem is to first identify the forces which you have done, secondly graph an x,y plane.
So based on that with reference to the right side of the picture you can conclude that (T - mg = ma) where a will be the acceleration of the block.
Secondly regarding the tension of the string being the same throughout, understand the concept of systems. Since the entire apparatus is connected by one string, it can be considered as a single system, if different parts of the string have different accelerations, the string would snap.Hence the acceleration of all the objects in the system, have the same acceleration.
