Objects that are quotient of two projective objects and cohomology in degree>1 1) What is an example of an abelian group which is not the quotient of two free abelian groups? 
For the abelian group $X$ for which this is true then for all Right exact functors F, i would have $L^iF(X)=0, \forall i >1$. So to make an example i started wondering which $F$ do not share this property, but it is an easy theorem that all the basic functors in this case has this property for abelian groups(Tor, Ext). So my next question is
2) What is an example of a right exact functor from $Ab$ to itself which has non trivial cohomology in degree greater than 1 (where nontrivial i mean non constantly trivial for all the objects)?
After having posed my self question 2) i started wondering in which way they could be related (that is, 1) always implies 2) so i'm wondering on the converse).
3) If a ring R has no examples in 2), can i conclude that every object admit a projective resolution that stops after two factors (that is every object is the quotient of two projective)? (The opposite being trivial as remarked). 
Thanks in advance!
Edit: The first question is trivial being the kernel always free over a PID.
Edit: given a reply down(that does not answer the question) let me specify that for being X a quotient of two projective i meant exactly that there is a short exact sequence $0 \to A \to B \to X \to 0$ with A,B projective. My question is: having such a property, for all X, is equivalent to have all cohomologies trivial for all right exact functors(from R-Mod to Ab) in degree 2 on? For wich rings this property holds?Is it equivalent to be a PID?
 A: A ring where every submodule of a projective module is projective (which is equivalent to your condition that every module is a quotient of a projective module by a projective submodule) is called "hereditary" ("left-hereditary" or "right hereditary" if the ring is non-commutative and you're considering left or right modules).
A commutative integral domain is hereditary if and only if it is a Dedekind domain, and there are many other examples that are non-commutative or not domains.
It is indeed true that being hereditary is equivalent to the fact that the second left derived functor of any right exact functor vanishes. A more standard fact is that $R$ being hereditary is equivalent to the vanishing of $\operatorname{Ext}^2_R$. So if $R$ is not hereditary then $\operatorname{Ext}^2_R(X,Y)\neq0$ for some modules $X$ and $Y$. Let $I$ be any injective abelian group such that $\operatorname{Hom}_{\mathbb{Z}}\left(\operatorname{Ext}^2_R(X,Y),I\right)\neq0$ ($I=\mathbb{Q}/\mathbb{Z}$ will do). Then $F=\operatorname{Hom}_{\mathbb{Z}}\left(\operatorname{Hom}_R(-,Y),I\right)$ is a right exact functor whose second left derived functor is $\operatorname{Hom}_{\mathbb{Z}}\left(\operatorname{Ext}^2_R(-,Y),I\right)$.
A: Every module over every ring is a quotient of two free modules: take the free module on a set of generators and the free module on a set of relations. What is not true in general, and is true for abelian groups, is the stronger statement that every module has a free resolution of length $2$. This requires that the map from the "relations" module to the "generators" module is injective. 
