Determine the elevation of the middle point of the curve.

I need the answer as soon as possible

1) An equal-tangent crest vertical curve has a 50 mi/h design speed. The initial grade is +3%. The high point is at station 33+37.43 and the PVT is at station 37+18.26. What is the elevation difference between the high point and the PVT?   2) A 1400-ft-long sag vertical curve (equal tangent) has a PVC at stateion 115+00 and elevation 750ft.  The intial grade is -3.5% and the final grade is +6.5%.  Determine the elevation and stationing of the low point, PVI, and PVT.   If you could show as many steps taken as possible that would be helpful in me actually understanding how to do the problem.  Thanks

Note : If you don't know the solution please leave it but don't give me wrong solution. A vertical curve joining entering grade of -4% with an exiting grade of - 7% begins at station 2+607 and is being designed with K value of 72. What is the station of the highest point on the curve . Code the station as the total distance from the origin, i.e., if the station is 1+100 code it as 1100.

3.7 An equal-tangent crest vertical curve is designed for 65 mi/h. The initial grade is +3.4% and the final grade is negative. What is the elevation difference between the PVC and the high point of the curve?