In Lecture Notes in Algebraic Topology, Davis & Kirk, it is written:
Proposition $\mathbf{2.4.}\,\,$ Let $R$ be a commutative ring and $a\in R$ a non-zero divisor (i.e. $ab=0$ implies $b=0$). Let $M$ be an $R$-module. Let $M/a=M/aM$ and $_aM=\{m\in M|am=0\}$. Then
$$ \begin{align} &1.\,\, R/a\otimes M \cong M/a, && 3.\,\,\operatorname{Hom}(R/a,M) \cong\ _aM, \\ &2. \,\,\operatorname{Tor}_1(R/a,M)\cong\ _aM, && 4.\,\,\operatorname{Ext}^1(R/a,M)\cong M/a. \\ \end{align} $$ Proof. Since $a$ is not a divisor of zero, there is a short exact sequence $$ 0\to R\xrightarrow[]{\times a}R\to R/A\to 0. $$ Apply the functors -$\otimes M$ and $\operatorname{Hom}(-,M)$ to the above short exact sequence. By the axioms we have exact sequences $$ 0\to\operatorname{Tor}_1(R/a,M)\to R\otimes M\to R\otimes M\to R/a\otimes M\to 0\text{$\,\,\,$ and} \\\,\\ 0\to\operatorname{Hom}(R/a,M)\to\operatorname{Hom}(R,M)\to\operatorname{Hom}(R,M)\to\operatorname{Ext}^1(R/a,M)\to 0. $$ The middle maps in the exact sequence above can be identified with $$ M\xrightarrow[]{\times a}M, $$ which has kernel $_aM$ and cokernel $M/a.\color{white}{\tag{$\color{black}{\square}$}}$
Question 1: Doesn't this claim work for arbitrary $a\in R$?
Question 2: Is there any nice formula for $\mathrm{Ext}^1_R(R/I,M)$ and $\mathrm{Tor}_1^R(R/I,M)$ given any ideal $I\unlhd R$?