Two blocks of masses 10g and …

For two blocks of masses fastened to the ends of a cord which passes over a frictionless pulley we have as per Newton's Second law :
motion for 30g or 0.03 kg mass will be , 0.03 g - T = 0.03 a ---------(1)
motion for 10g or 0.01 kg mass will be , T - 0.01 g = 0.01 a ---------(2)

Now the answer for the third part of the question can be obtained by solving (1) & (2) and putting g = 9.8 m/s^2

a = 4.9 m/s^2 and T = 0.147 N

for the first and second part of the question the eqn (2) has to be modified as under to accommodate minimum force, t, say to be applied . 't' will be applied downward to keep the 0.01kg mass from rising up
Also now the system will be at rest and so a = 0

The eqns will be now as under :

0.03 g - T = 0.03 x0 = 0 ----------------(3)
T - 0.01 g - t = 0.01x0 = 0 --------------(4)

solving (3) & (4) we get ,

t = min force to be applied = 0.196 N
T = tension = 0.294 N