a) the ball's initial velocity is going to be the same as its velocity when it returns to its starting position if you dropped it from 1.54m. For this I would simply use v^2 = 2ax, where "a" is the acceleration of gravity, 9.8m/s^2.
b) Use the numbers you now have and just plug them into x = 1/2at^2 and solve for "t".
c) It takes the ball the exact same amount of time to fall to its initial position as it took to rise to its maximum position, so this will just be double whatever your answer to question (b) was.
Honestly man, all you have to do is look up your equations for distance, velocity, etc. I would guess if your homework involves this sort of thing, you've probably been given them. All you have to do is put in your numbers and plug and chug. It's simple algebra.