Exercise 2.4.3., pg 47 If $0 \rightarrow M \rightarrow P \rightarrow A \rightarrow 0$ is exact with $P$ projective (or $F$-acyclic), then $L_iF(A) \cong L_{i-1}F(M)$ for $i \ge 2$ and that $L_1F(A)$ is the kernel of $F(M) \rightarrow F(P)$
I could do the first part: simply construction a resolution $P_2 \rightarrow P_1 \rightarrow M$ and connect it to $P$, via $P_0 \rightarrow M \rightarrow P$.
I have trouble showing the second. It would really help if someone spells out the details.
I will describe the whole argument, because your approach doesn't quite sound like dimension shifting to me as I think of it.
Let $0\to M\to P\to A\to0$ be a short exact sequence, $F$ a right-exact functor, and $P$ a projective (or at least $F$-acyclic) module. Then we get the corresponding long exact sequence
\begin{align*} \cdots&&\to &&L_{2}F(M)\to &&L_{2}F(P)\to &&L_{2}F(A)&&\\ &&\to &&L_{1}F(M)\to &&L_{1}F(P)\to &&L_{1}F(A)&&\\ &&\to &&F(M)\to &&F(P)\to &&F(A)&&\to0. \end{align*}
Since $P$ is $F$-acyclic, $L_{i}F(P)=0$ for $i\neq0$. Then we have
\begin{align*} \cdots&&\to &&L_{2}F(M)\to &&0\to &&L_{2}F(A)&&\\ &&\to &&L_{1}F(M)\to &&0\to &&L_{1}F(A)&&\\ &&\to &&F(M)\to &&F(P)\to &&F(A)&&\to0. \end{align*}
and since the sequence is exact, $L_{i}F(A)\cong L_{i-1}F(M)$ for $i\geq2$, as desired. By the same long exact sequence, we have
\begin{align*} 0\to L_{1}F(A)\to F(M)\to F(P), \end{align*}
so $L_{1}F(A)$ is the kernel of $F(M)\to F(P)$.