De Moivre's Theorem and Nth Roots of Unity Question

I've been having difficulty with the second and third parts of this question, any help would be appreciated.

Consider the complex number z = cis ([2Pi(k)]/5) for any integer k such that z does not equal 1.

a) Show that z^n + z^-n = 2cos([2Pi(k)(n)]/5) for any integer n.

I was able to verify this result, the next part is where I'm a bit a confused:

b) Show that z^5 = 1. Hence, show that 1+z+z^2+z^3+z^4 = 0

Using the earlier result, I substituted z^5 in for 1 so that

z^n + z^5/z^n = 2cos([2Pi(k)(n)]/5)

substituting values for n, I was able to come up with the pairs:

z^4 + z^1 and z^2 +z^3

I'm now stuck, although placing the respective values given by the earlier result into my calculator gives me the desired outcome, I am looking for a way to prove the result purely through algebra.

I also noticed that the given equation is in a geometric progression with initial term 1 and common ratio z:

Using the sum formula:

S = 1(1-r^n)/(1-r)

Sum of first 5 terms :

= 1(1-z^5)/(1-z)

=1(0)/1-z

=0

This gives the desired outcome, but I am looking to get the same result using the earlier parts of the question - I haven't had any luck.

c) Find the value of b, given that:

(z+z^-n)^2 + (z^2 + z^-2)^2 = b

I have tried using the earlier result and then using trig identities to try to solve for b, but I can't seem to solve the equation.

Sorry if this is too many questions, but I am also a bit confused about the following question:

If n is a positive integer, show that (1+i)^n + (1-i)^n = 2^((n/2)+1) * cos([Pi(n)]/4)

I was able to simplify this to 2cos([Pi(n)]/4) but I do not really understand how the exponent ((n/2)+1) comes about

Thanks in advance