Tension/Acceleration Problem

I am a bit confused as to how to tackle a physics problem containing multiple tensions, as in the following (I tried as best I could to represent what appears on my homework sheet; but if you're wondering, there's a platform on which the 1kg and 4kg blocks are moving, the O is the pulley, Tension1 is the rope between the 3kg and 4kg blocks, and Tension 2 is that between the 4kg and 1kg blocks):

O– – –4kg——1kg
I
I
I
3kg

Prompt: If the system rests on a frictionless surface, find the acceleration of the system and the tension in each segment of rope (as indicated by the solid and doted lines).

So I think acceleration (A) would be 2.45 m/s^2 (since, by the equation F=MA, 9.8+39.2–29.4 = (1+4+3)A), but I'm not sure how to obtain tension (I tried doing (3x9.8)–T1 = 3(2.45) and found T1 = 22.05, but by this logic there would be one tension for each block, not rope segments).

Thanks for the help!

The total force is 3*g = 29.4N

The total mass accelerated is 3+4+1 = 8kg

Since all the blocks are tied together, they have the same acceleration

F = m*a => F/m = a = 3g/8 = 3.675m/s^2 <--------------------

If the 4kg and 1kg blocks were stationary,

T1 would be 3*g = 3*9.8 = 29.4 but

Since the 4kg+1kg blocks are accelerating

T1 = ma = [4+1]*3g/8 = 5*3.675 = 18.375N <--------------------

If the 1kg block were stationary,

the tension would be 3*g but since the 1kg block is

accelerating T2 = ma = 1*3g/8 = 3.675N <--------------------

f=3*9.8=29.4
a=29.4/(3+4+1)=3.675(they all accelerate the same, by the force of the 3kg)
t1=a*(4+1)=18.375(it is the pull from the 3kg on the 5 kg)
t2=a*1=3.675(the force needed to accelerate the 1kg as all of them)