Part 1: correct
Part 2:
Multiply by 100:
Which has solutions (thanks to WolframAlpha, or you can complete the square):
so at (approx) x = 0.15ft and x = 1.35ft the frog will be 3.6ft above the ground.
Part 3: For this, we can use Calculus (take dh/dx, set to zero, solve for x) or we can note that for a parabola the max/min will be on the axis of symmetry which is at x = -b/(2a):
ft
[ Calculus method: dh/dx = -100x + 75
-100x + 75 = 0 --> x = -75/-100 = 0.75ft ]
Part 4: Plug in x = 0.75 into h(x):
ft (approx)
Some/most tutors will ignore your post if you post multiple problems. I can understand why you did it :-)