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.
Can you express the set of primitive $n$-th roots of unity in terms of $n$ using $\gcd$? Once you have that, it should not be hard to find the two desired sets and their intersection.
[Edit: Here is the way I had in mind, which uses more than necessary.]
(We actually know what the set of primitive $n$-th roots of unity are, if we have just one of them. I will give a proof here, but the result is well-known.)
If $r$ is a primitive $n$-th root of unity,
Let $f(x) = x^n-1$
Let $S = \{ r^k : k \in \{1,2,...,n\} \}$
$S$ contains only roots of $f$ because $(r^k)^n = (r^n)^k = 1^k = 1$
$S$ contains exactly $n$ elements otherwise:
$r^{k-m} = 1$ for some distinct $k,m \in \{1,2,...,n\}$
$r$ is not a primitive $n$-th root of unity because $|k-m|<n$
$\Rightarrow\Leftarrow$
Therefore $S$ is all the roots of $f$ because $f$ has at most $\deg(f) = n$ roots
The primitive $n$-th roots of unity are $\{ r^k : k \in \{1,2,...,n\} \wedge gcd(k,n) = 1 \}$
(Note that some fields do not have primitive roots, such as $\mathbb{R}$ does not have a primitive cube-root of unity, and $\overline{\mathbb{F}_2}$ does not have a primitive square-root of unity. But the algebraic closure of any field of characteristic $0$ has $\phi(n)$ primitive $n$-th roots of unity.)
Now here is an alternative method that doesn't make use of the above information at all:
If $r$ is both a $42$nd and $70$th primitive root of unity,
$r^{14} = r^{gcd(42,70)} = 1$ and hence a contradiction
(In general, the sets of primitive $n$-th roots of unity are disjoint and partition the set of all roots of unity.)
Best Answer
Primitivity means that no positive power of $\zeta=e^{2\pi i k/n}$ less than $n$ will achieve unity. If $k$ is not coprime to $n$ and $\gcd(k,n)=m$, then observe $\zeta^{(n/m)}=e^{2\pi i (k/m)}=1$ but $n/m<n$ if $m>1$.