Let $R$ be a commutative unital ring and $r\in R$. Let $A$ and $B$ be $R$-modules. Does $rA=0$ or $rB=0$ imply $r\operatorname{Ext}^n_R(A,B)=0$ for all $n\in\mathbb{N}$?
For $n=0$ it holds, but I'm not sure about $n\geq1$.
I was thinking about the exact sequences $$0\to rM\to M\to M/rM\to0,$$ $$0\to r\operatorname{Ext}\to \operatorname{Ext}\to\operatorname{Ext}/r\operatorname{Ext}\to0,$$ $$0\to \operatorname{Ann} M\to R\to R/\operatorname{Ann} M\to0,$$ but they don't seem to tell us anything of use from the long exact Ext sequence.
I've also tried to prove it via the definition of Ext, but so far unsuccessfully. I don't know what happens with $r\cdot$ after Hom is applied to a projective resolution.
As Darij Grinberg points out, we only have to prove that $\mathrm{Ext}^n$ is $R$-linear in both variables.
For this we have to recall how the action on morphisms is defined: If $f : A' \to A$ is a homomorphism, and $P^* \to A$ and $P'^* \to A'$ are projective resolutions, there is a (unique up to homotopy) homomorphism of complexes $\overline{f} : P'^* \to P^*$ such that
$$\begin{array}{c} P'^* & \rightarrow & A' \\ \downarrow && \downarrow \\ P^* & \rightarrow & A \end{array}$$
commutes. It induces a homomorphism of complexes $\hom(P^*,B) \to \hom(P'^*,B)$, and then a homomorphism on cohomology $\mathrm{Ext}^n(f,B) : \mathrm{Ext}^n(A,B) \to \mathrm{Ext}^n(A',B)$.
Now, if $r \in R$, then we can multiply the vertical maps in the diagram above with $r$, and see that $r \overline{f}$ is a choice for $\overline{rf}$. The rest of the construction is also compatible with multiplication with $r$. Hence, $\mathrm{Ext}^n(rf,B) = r \mathrm{Ext}^n(f,B)$.
The proof for the other variable is identical. And for $\mathrm{Tor}$ the same proof works.
Since you ask "Is this implicit in the construction of Ext and Tor when R is commutative?", let me add what is really going on here: There are two Ext functors! The first one, which also exists in the non-commutative case, in fact in arbitrary abelian categories with enough injectives, is $\mathrm{Ext}^*(A,-) : \mathsf{Mod}(R) \to \mathsf{Ab}$, the right derived functor of $\hom(A,-) : \mathsf{Mod}(R) \to \mathsf{Ab}$. The second is what I would denote by $\underline{\mathrm{Ext}}^*(A,-)$, the right derived functor of $\underline{\hom}(A,-) : \mathsf{Mod}(R) \to \mathsf{Mod}(R)$. If $U$ denotes the forgetful functor $\mathsf{Mod}(R) \to \mathsf{Ab}$, then we have an isomorphism $U \circ \underline{\mathrm{Ext}}^*(A,-) \cong \mathrm{Ext}^*(A,-)$. You can prove this as follows: They are universal exact $\delta$-functors and agree for $*=0$.
The distinction between $\mathrm{Ext}$ and $\underline{\mathrm{Ext}}$ becomes more important for the category of modules on some ringed space $X$. They are connected by a "local-global" spectral sequence $H^p(X,\underline{\mathrm{Ext}}^q(A,B)) \to \mathrm{Ext}^{p+q}(A,B)$. If $X$ is just a point, the spectral sequence degenerates to the isomorphism above.