since the blocks are all connected in one way or another, i think that the acceleration of each would be the same. I am really not sure though, but i think it may be a=M3g/(M1+M2+M3). I'm not 100% sure, but i think that is correct
for part a, you don't have to draw a force for the contact. it is assumed that friction is negligible and won't affect it.
for part c, the tension will remain the same seeing as though the length of the string and the masses do not change. Therefore, you are correct and the "initial T" will...
for part b, x: (M1+M2+M3)(a)=F_x. solve for a. Since you want it to remain stationary, a_y=0=T-M3g. (M2)a_x=T. So then you have the acceleration you solved for multiplied by M2 and that will equal M3g. Do algebra to get F= M3g(M3+M2+M1)/M2.