Just a little help.
If $\zeta_k=e^\frac{2\pi ik}{n}$ is a primitive $n^\text{th}$ root of unity for $n\in\mathbb{Z}^+$, then $k\in\mathbb{Z}$ is relatively prime to $n$. Furthermore, $\zeta_j=\zeta_k\iff j\equiv k\pmod n$. So the set of all primitive $n^\text{th}$ roots of unity is parametrized by the set $K$ of all $1\le k<n$ relatively prime to $n$, i.e. the reduced residue system modulo $n$, which can be identified with $(\mathbb{Z}/n\mathbb{Z})^*$.
Now $s(n,p)=\zeta_k^p=e^\frac{2\pi ikp}{n}=\cos\frac{2\pi kp}{n}+i\sin\frac{2\pi kp}{n}$
and $\Re\,\sum\zeta_k^p=\sum\cos\frac{2\pi kp}{n}$, so we might hope that the full complex sum is zero or not zero under the respective conditions.
This will depend on what happens to the residues $k$ modulo $n$ when they are multiplied by $p$. For $p$ relatively prime to $n$, they get permuted ($pK=K$), so that the sum is zero (in the complex plane, both the real and imaginary parts). If, however, $d=\gcd(p,n)>1$, then $\zeta_k^p$ is a primitive root of order $\frac{n}{d}$. So how do we get a complex sum of zero from primitive roots? One way is to sum all the vertices of the regular $n$-gon, i.e. all $n^\text{th}$ roots of unity: $\sum_{j=0}^{n-1}\zeta_j=0$. However, these can be regrouped by their order: $\sum_{d|n}s(d,1)=0$, in terms of the sums $s(n,p)$ defined above. What is $\sum_{d|n}s(d,p)$?
Let $r=\href{http://en.wikipedia.org/wiki/Radical_of_an_integer}{\text{rad}}\,n$
$=\prod_{q\mid n}q$ be the product of all primes $q$ dividing $n$ (also called the radical or squarefree kernel of $n$).
What does it tell us when the positive (but not necessarily prime) integer $p$ is less than $\frac{n}r$? Or equal to $\frac{n}r$? What does that tell us about $\frac{n}p$ in relation to $r$? It must be less than $r$ and therefore relatively prime to at least one prime $q$ dividing $r$ and $n$.
You may also find the Möbius inversion formula helpful; your $\mu$ is $\mu(r)$, where my $\mu$ is the Möbius function.
First, there are at most $n$ $n$th roots of unity, because $x^n-1$ can have at most $n$ roots (as a consequence of the Factor Theorem applied in $\mathbb{C}$).
Second, if $\omega$ is an $n$th root of unity, that means that $\omega^n = 1$. But then, for any integer $k$, we have
$$(\omega^k)^n = \omega^{kn} = (\omega^n)^k = 1^k = 1,$$
so $\omega^k$ is also an $n$th root of unity.
So now the question is which ones are different? If $\omega$ is such that $\omega^n=1$ but $\omega^{\ell}\neq 1$ for any $0\lt \ell\lt n$, then $\omega^r=\omega^s$ if and only if $\omega^{r-s}=1$, if and only if $n|r-s$: indeed, using the division algorithm, we can write $r-s$ as $qn + t$, with $0\leq t\lt n$ (division with remainder). So then
$$1=\omega^{r-s} = \omega^{qn+t} = \omega^{qn}\omega^t = (\omega^n)^q\omega^t = 1^q\omega^t = \omega^t.$$
But we are assuming that $\omega^t\neq 1$ if $0\lt t\lt n$; since $0\leq t\lt n$ and $\omega^t=1$, the only possibility left is that $t=0$; that is, that $n|r-s$.
Thus, if $\omega^{\ell}\neq 1$ for $0\lt \ell\lt n$ and $\omega^n=1$, then $\omega^r=\omega^s$ if and only if $n|r-s$, which is the same as saying $r\equiv s\pmod{n}$.
So it turns out that $\omega^0$, $\omega^1$, $\omega^2,\ldots,\omega^{n-1}$ are all different (take any two of the different exponents: the difference is not a multiple of $n$), and they are all roots of $x^n-1$, and so they are all the roots of $x^n-1$.
So it all comes down to finding an $\omega$ with the property that $\omega^k\neq 1$ for $0\lt k\lt n$, but $\omega^n=1$. And
$$\large\omega = e^{2\pi i/n}$$
has that property.
Best Answer
$$\sum_{k=1}^n\cos^2(2\pi k/n)=\frac{1}{2}\sum_{k=1}^n(\cos(4\pi k/n)+1)=\frac{n}{2}$$ since $\sum_{k=1}^n\cos(4\pi k/n)=\mathrm{Re}\sum_{k=1}^ne^{4\pi ki/n}=0$.