Please help with cumulative distribution function question

Using the following cumulative distribution function:

Fx(x) = {
0 if x < 1
1/16 (x-1)^2 if 1 ≤ x ≤ 5
1 if x > 5

1) Find the probability density function, fx(x), of X.

2) Calculate E(X), the expected value of X.

3) Let A be the event {X > 3}.
a) What is P(X > 3)?
b) What is Fx|a(x) = P(X ≤ x | X > 3) for -Infinity < x < Infinity?
(Hint: consider separately the cases x < 3, 3 < 5, x > 5.)
c) What is Fx|a, the probability density function of X given X > 3?
d) What is E(X | X > 3)?

Few questions here to be answered, please anyone / experts, your helps are much appreciated here.

1) Just take the derivative of Fx:
fx(x) = (1/8)(x - 1) if 1 < x < 5, and 0 otherwise.

2) E(X) = ∫(x = 1 to 5) x * (1/8)(x - 1) dx
...........= ∫(x = 1 to 5) (1/8)(x^2 - x) dx
...........= (1/8)(x^3/3 - x^2/2) {for x = 1 to 5}
...........= 11/3.

3a) P(X > 3)
= ∫(x = 3 to 5) (1/8)(x - 1) dx
= (1/8)(x^2/2 - x) {for x = 3 to 5}
= 3/4.
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b) Fx|a(x)
= P(X ≤ x | X > 3)
= P(X ≤ x and X > 3) / P(X > 3)

If x < 3, then Fx|a(x) = 0.

If 3 < x < 5, then we have
Fx|a(x) = P(3 < X ≤ x) / P(X > 3)
...........= [∫(3 to x) (1/8)(x - 1) dx] / (3/4)
...........= [(1/8)(x^2/2 - x) - 3/16] / (3/4).

If x > 5, then Fx|a(x) = [(1/8)(5^2/2 - 5) - 3/16] / (3/4) = 1.
-----------
c) Differentiate the result from part b:
Fx|a = [(1/8)(x - 1)] / (3/4) = (1/6)(x - 1) for x in (3, 5) and 0 otherwise.
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d) ∫(x = 3 to 5) x * (1/6)(x - 1) dx
= ∫(x = 3 to 5) (1/6)(x^2 - x) dx
= (1/6)(x^3/3 - x^2/2) {for x = 3 to 5}
= 37/9.

I hope this helps!