For question 2, note that the total mass is 2m and this is undergoing a constant acceleration due to gravity. The constant acceleration equations allow the velocity, v, at any time to be written as: v = u + at, where u is initial velocity, a is acceleration (due to gravity here) and t is time interval.
For question 3, note that the 700g mass is being pulled down by gravity as is the 500g mass. However, because they are on opposite sides of a pulley, there will be a net force of (0.7 - 0.5)*9.8 N pulling the 700g mass downwards and the 500g mass upwards. Find the time it takes the 700g mass to reach the floor using s = ut + (1/2)at2, where s is the distance (20cm). Find the velocity of both 700g and 500g (the same for both, of course) from v1 = u + at. This will act as the initial upward velocity for the 500g mass which will carry on rising against gravity, but slowing until it comes to a stop.
Use 0 = v12 + 2as to find the extra distance, s, travelled by the 500g mass. Don't forget to add on the 20cm to get the total height, and to use -9.8m/s2 for gravitational acceleration in this part (as the mass is travelling in the opposite direction from that in which gravity is acting).
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