On representations of a nonabelian group of order $pq$ 
Let $p,q$ primes number s.t. $p>q$ and let $G$ a non abelian group of order $pq$.
1) Determine all degree of irreducible representation
2) Show that $|[G,G]|=p$ (where $[G,G]=\left<ghg^{-1}h^{-1}\mid g,h\in G\right>$)
3) Show that $q$ divide $p-1$ and that $G$ has $q+\frac{p-1}{q}$ conjugacy classes.

My answers
1) There is of course the trivial representation of degree 1. I know that if there is $k$ irreducible representation, then $1+d_2^2+\cdot +d_k^2=pq$ where $d_i$ it the degree of the $i-$th representation, but I don't know how to apply it here. 
2) Since $[G,G]$ is a subgroup of $G$ and that $G$ is not abelian, $|[G,G]|\neq 1$ and thus $|[G,G]|\in\{p,q,pq\}$. Now I know that a group of order $pq$ with $p$ and $q$ prime is resoluble, and thus, if $|[G,G]|=pq,$ the group $G$ wouldn't have any subgroup $H$ s.t. $G/H$ abelian what would be a contradiction with the fact that it's resoluble (isn't it ?). And thus, $|[G,G]|\in\{p,q\}$. Now, each group of prime order is abelian, then $[G,G]$ is abelian. But how can I choos between $p$ and $q$ ?
3) I have no idea.
 A: For 2, remark that by using Sylow theorem, the number of Sylow p-group divides q and is equal to 1 mod p. If this number is q, then it is impossible since q is not 1 mod p since $q<p$. You deduce that the P p Sylow subgroup is normal.
You have [G,G] is normal it is different of G since it's image by the quotient map G to G/[G,G] is triavial. Its cardinal is not 1 since G is not abelian. If it cardinal was q, then the intersection of [G,G] and P is trivial since their orders are relatively prime. This fact implies also that P commutes with [G,G] since the commutator of am element of P and [G,G] is contained in their intersection since they are normal. This implies that G is the product of P and [G,G]  and henceforth commutative. Contradiction.
For 3 consider $n_q$ the number of Sylow q-subgroups, you have $n_q$ divides p and $n_q$ is equal to 1 mod q. $n_q$ can't be 1 since the q Sylow subgroup Q will be normal and as above we remark that the commutator of an element of Pan Q is in their intersection so G is commutative, you deduce that G is commutative, contradiction. Thus $n_q=p$ and $p=1$ mod q. This implies that q divides $p-1$
A: For 1), Let $\chi_1,...,\chi_n$ the irreducible characters. You know that $$\chi_1^2(1)+...+\chi_n^2(1)=|G|=pq.$$
You also know that $\chi_i(1)$ must divide $pq$ for all $i$, and thus $\chi_i(1)\in \{1,p,q,pq\}$ for all $i$. Since $(pq)^2>pq$, and $p^2>pq$ you in fact get $\chi_i(1)\in \{1,q\}$, what conclude the proof. 
