Every profinite group is a quotient of a profinite free group by a normal subgroup that is free profinite? It is well known that any group is a quotient a free group by a normal subgroup that is free. More precisely if $G$ is a group the exists a short exact sequence of groups 
$$1\rightarrow F^{'}\rightarrow F\rightarrow G\rightarrow 1 $$
where $F^{'}$ and $F$ are free groups. 

Q1:
      Is any profinite group a quotient of a free profinite group by a normal subgroup that is free profinite?

Assuming that the answer to the first question is negative

Q2:
      what can we say about a profinite group if initially we
      know that it is a quotient of a free profinite group by a normal subgroup that is free profinite?  

By the second question I do mean if such profinite group has some cohomological properties. 
 A: We answer Q1 in affirmative, using the mentioned theory of Oleg V. Melnikov. We refer to Fried, Jarden, Field Arithmetic, 3rd edition as [FJ].
First some notation.
For a group $F$ we denote by $|F|$ its cardinality.
Let $F$ be a free profinite group of an infinite rank $m\ge|G|$.
There is an epimorphism $F \to G$. Let $N$ be its kernel, then $G \cong F/N$.
Let $S$ be a finite simple group. Then $r_F(S)=m$, that is,
the largest quotient of $F$ which is the product of copies of $S$,
is isomorphic to $S^m$ [FJ, Lemma25.7.1].
Notice that $|S^m|=2^m$.
The quotient map $\pi\colon F \to S^m$ maps $N$ onto a closed normal subgroup of $S^m$, hence $\pi(N)\cong S^\kappa$ for some cardinality $\kappa\le m$.
Now, $|\pi(F)/\pi(N)|\cdot|\pi(N)|=|\pi(F)|$,
that is,
$|\pi(F)/\pi(N)|\cdot 2^\kappa=2^m$,
but $|\pi(F)/\pi(N)|\le |F/N|=|G|\le m$,
so $\kappa= m$.
Hence, by [FJ, Theorem 25.7.3(b)], $N\cong F$.
Another version of the proof:
Instead of $m\ge |G|$ assume just $m\ge\text{rank}\;G$.
Then there is still an epimorphism $F \to G$.
In fact, we can take it to be the composition of epimorphisms
$F\to F\times G \to G$; then its kernel $N$ has $F$ as a quotient.
Then [FJ, Lemma 24.9.2(a)] gives $r_N(S)\ge r_F(S) = m$,
so $r_N(S)=m$, for every finite simple group $S$.
Hence, by [FJ, Theorem 25.7.3(b)], $N\cong F$.
