Our fill depths are y values, so you can plug and chug on your own. y = y(x) (y is a function of x)
y(-500), y(-400), y(-300), etc. You should have no problem plugging x values into our function.
_________
Part C is going to have you take not of that issue I mentioned earlier. But lets just do this using calculus since this is a calculus class.
Extrema occurs when the derivative is equal to zero.
y = 0.000075 x^2 - 0.015 x
dy/dx = 0.00015 x -0.015
Set this bad, boy to zero and solve....
0 = 0.00015 x -0.015
x = 100.
That is a nice number, so thankfully the maker of the question is kind. I really hope this helped, if you need help with part b just ask, but I think once you have part A, B is cake.
[Edit]
After reading Deepak's argument carefully I see he did make one assumption that I would not have made. By saying 0.09 * 500 (though remember it should be -0.09) he is assuming the slope is constant, which clearly it is not, in the case of a parabola. It also, as you pointed out, eliminated all the calculus from the problem, but it is still clever and yielded a pretty close to correct curve too.