Show that $\text{Ext}^n (X,Y) = 0$ for all $n \ge 2$ and give an example in which $X$ and $Y$ be two Abelian groups but $\text{Ext} (X,Y) \ne 0$ 
Problems: Let $X$ and $Y$ be two Abelian groups. Show that $\text{Ext}^n (X,Y) = 0$ for all $n \ge 2$. Give an example in which $X$ and $Y$ be two Abelian groups but $\text{Ext} (X,Y) \ne 0$.

My attempt: Consider the exact sequence
$$K \colon 0 \rightarrow A \rightarrow F \rightarrow X \rightarrow 0$$
$F$ be a free Abelian group, implies $A$ is a free group. Hence, $K$ be a projective resolution of $X$. We have 
$$\text{Hom}(K,Y) \colon 0 \rightarrow \text{Hom}(A,Y) \rightarrow \text{Hom}(F,Y) \rightarrow \text{Hom}(X,Y) \rightarrow 0$$
is also an exact sequence.
Could you give me some suggestion to continue the proof? Thank all!
 A: You are on the right line by finding a projective resolution of $X$, since $\mathrm{Hom}(-,Y)$ is a contravariant left exact functor, and so to compute the right derived functors of it we need to apply it to a projective resolution of $X$. Really, the projective resolution of $X$ is the chain complex
$$
0\leftarrow F \leftarrow A \leftarrow 0
$$
where $F$ is in degree $0$. Then, when we apply $\mathrm{Hom}(-,Y)$ to this projective resolution, we get a cochain complex 
$$
0 \rightarrow \mathrm{Hom}(F,Y) \rightarrow \mathrm{Hom}(A,Y) \rightarrow 0
$$
where $\mathrm{Hom}(F,Y)$ is in degree $0$. Now the group $\mathrm{Ext}^n(X,Y)$ is the $n$-th cohomology group of this complex. But since the complex is $0$ in degrees $\ge2$, that must mean $\mathrm{Ext}^n(X,Y)=0$ for all $n\ge 2$.
The exact sequence
$$
0\rightarrow\mathrm{Hom}(X,Y) \rightarrow \mathrm{Hom}(F,Y) \rightarrow \mathrm{Hom}(A,Y) 
$$
shows, as always, that $\mathrm{Ext}^0(X,Y)=\mathrm{Hom}(X,Y)$. So the only interesting ext group to calculate in this case is $\mathrm{Ext}^1$.
