It looks like the maximum height is at t=2. This is an assumption based on the height being 5 at t=1 and t=3, and assuming that there are no weird complexities between these times.
The average velocity at the maximum height is given by
v = ∆s / ∆t
where ∆t = 2 and ∆s is the distance between the heights at t=2 and t=0.
Instantaneous velocity is given by the slope of the tangent line, which is given for questions b-d.
If you ignore friction, the launch velocity and the final velocity should be the same, since height is the same at the start and the end.