Prove that if a group is nilpotent , then its quotient with its Frattini subgroup is abelian I know that :
1) Nilpotent group is solvable.  
2) Subgroup of a solvable group is solvable.
3) Solvable and simple group is abelian.
Now I should use these facts to prove it.
 A: If I understand you correctly you wish to show that if $G$ is nilpotent then the quotient $G / \Phi(G)$ is abelian. I don't know how to solve it using just the facts you stated but using (1) that every maximal subgroup of a nilpotent group is normal and (2) quotients of solvable (nilpotent) groups are solvable (nilpotent) (either of these two works). 
To show this it suffices to show that the commutator subgroup $G'$ is contained in the Frattini $\Phi(G)$. (*) By definition $\Phi(G)$ is the intersection of all the maximal subgroups of $G$. It is a fact about nilpotent groups that all maximal subgroups are normal. So let $M$ be a maximal subgroup of $G$. Then since $M$ is maximal $G/M$ is a simple group. A quotient of a nilpotent group is nilpotent, hence $G/M$ is nilpotent. So $G/M$ is simple and nilpotent, thus abelian.
It is easy to see that if $G/M$ is abelian then $G' \leq M$ (just show directly that all the commutators are in $M$). Hence for any maximal subgroup $M$, $G'\leq M$, which implies that $G' \leq \Phi(M)$.
(*) It is easy to see that if $G' \leq N$ for any normal subgroup $N$ of $G$, then $G / N$ is abelian (just show that all the commutators are trivial, using the fact that $G' \leq N$). 
