The definition of $\operatorname{Tor}$ I am using is: Let $K \to F\to A\to 0$ be a free resolution of $A$ and $B$ an abelian group, then $\operatorname{Tor}(A,B) := \ker (f \otimes 1_B)$ if $f$ is the map $f\colon F \to A$.
What is the free resolution in this case? Is it $$\mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}_m \to 0$$ where the first map is multiplication by $m$ and the second is projection?
If so, how does one proceed?
You have the free resolution $\dots\to 0\to \mathbb Z\xrightarrow{m}\mathbb Z\to\mathbb Z/m\to 0$.
Now, ignore the last term and tensor with $\mathbb Z/n$:
$$\dots\to0\to \mathbb Z\otimes\mathbb Z/n\xrightarrow{m\otimes 1_{\mathbb Z/n}}\mathbb Z\otimes\mathbb Z/n\to 0\to\cdots,$$ which is isomorphic (as an exact sequence) to
$$\dots\to0\to\mathbb Z/n\xrightarrow{m}\mathbb Z/n\to 0\to\cdots.$$ Now $\mathrm{Tor}^\mathbb Z_1(\mathbb Z/m,\mathbb Z/n)$ is the homology at the first term, which is
$$\ker(m)=\frac n{\gcd(m,n)}\mathbb Z/n\cong\mathbb Z/\gcd(m,n).$$
