If you draw free body diagrams for each mass, you will obtain these three equations. (Sorry I can not show you the diagrams) For each diagram, you have to apply the Newton's 2nd law.
1) for m3: 3g  T3 =3a (T3 is the tension between m2 and m3, "g" is gravitational acceleration = 9.8 m/s^{2 } and "a" is the acceleration of the three masses which falling to the m2 and m3 side)
2) for m2: 4.2g + T3 T2 = 4.2g (T2 is the tension between m1 and m2)
3) for m1: T2  5.5g = 5.5a (Please be careful when you write this eaquation, you have to consider the direction of acceleration. Put the direction of acceleration as a positive direction)
If you substitute 9.8 for "g" and solve the 3 linear first order equations then you can obtain the answers.
(a) acceleration a = 1.312 m/s^{2} for each mass
(b) T2 = 61.116 N and T3 = 25.466 N
The problem wants up to 2nd decimal position so you have to round the given answers to fit the given significant numbers. If there are any required conditions, you have to fit that requirement related with the significant numbers.
I hope you can understand this question without diagrams. I feel sorry about the limitation of this answering box.
11/24/2012

Sung taee L.
Comments
Wow, I had a problem with the answer box, here, too... let me try to reorganize a bit...
You have to draw a force diagram to answer this question.
(T1)
^


O m1


v
(m1*g)