I am reading ''Algebraic number theory'' by Cassels and Fröhlich and in the chapter IV.4 it says the following:
If $\dots\to P_2\to P_1\to P_0\to\mathbb{Z}\to 0$ is a projective resolution and $G$ is an abelian group then the following sequence is almost exact (all homology grups are $0$ except for the first one): $$0\to \text{Hom}(P_0,G)\to \text{Hom}(P_1,G)\to \text{Hom}(P_2,G)\to\dots$$
The only reasoning for this that is written in the book is ''Because all the $P_i$ are free as abelian groups''.
I know that the $\text{Hom}(-,G)$ is a contravariant left exact functor and when $G$ is injective it is exact (not sure if this helps here) but I don't know much about how to approach this problem. I know I shoudn't post a question without any try, but here I am totally lost, so I will understand that you won't solve the question for me, but any small hint will be thanked.
A long exact sequence $$\dots \to A_{n+1} \xrightarrow{f_{n+1}} A_n \xrightarrow{f_n} A_{n-1} \to \dots$$ can be broken down as a bunch of short exact sequences $$0 \to \ker f_n \to A_n \to \operatorname{img} f_n \to 0$$ which stitch together via the equalities $\ker f_n = \operatorname{img} f_{n+1}$.
When we apply $\operatorname{Hom}({-},B)$ to the above short exact sequence, we get the long exact sequence $$0 \to \operatorname{Hom}(\operatorname{img} f_n, B) \to \operatorname{Hom}(A_n, B) \to \operatorname{Hom}(\ker f_n, B) \to \operatorname{Ext}^1(\operatorname{img} f_n, B) \to \operatorname{Ext}^1(A_n, B) \to \dots$$ If each $A_n$ is projective (as in your post), this yields an exact sequence $$0 \to \operatorname{Hom}(\operatorname{img} f_n, B) \to \operatorname{Hom}(A_n, B) \to \operatorname{Hom}(\ker f_n, B) \to \operatorname{Ext}^1(\operatorname{img} f_n, B) \to 0$$
In your situation, each $A_n$ is actually a free abelian group, and every subgroup of a free abelian group is free, so also each $\operatorname{img} f_n \leq A_{n-1}$ is free abelian. Thus this collapses to $$0 \to \operatorname{Hom}(\operatorname{img} f_n, B) \to \operatorname{Hom}(A_n, B) \to \operatorname{Hom}(\ker f_n, B) \to 0$$
Again since $\operatorname{img} f_{n+1} = \ker f_n$, these stitch together to give a long exact sequence $$\dots \to \operatorname{Hom}(A_{n-1},B) \to \operatorname{Hom}(A_n,B) \to \operatorname{Hom}(A_{n+1},B) \to \dots$$
To make this whole argument even simpler: in your situation, the short exact sequences $0 \to \ker f_n \to A_n \to \operatorname{img} f_n \to 0$ are all split, because all three objects are free.