Let $m,n$ be positive integers with $m\mid n$. I want to compute $$ \mathrm{Ext}_{\mathbb{Z}/n\mathbb{Z}}^i (\mathbb{Z}/m\mathbb{Z},\mathbb{Z}/m\mathbb{Z}) \qquad \mathrm{Tor}^{\mathbb{Z}/n\mathbb{Z}}_i (\mathbb{Z}/m\mathbb{Z},\mathbb{Z}/m\mathbb{Z}) $$ for $i \ge 0$. At first, I thought that the way to do this was to find a projective resolution of $\mathbb{Z}/m\mathbb{Z}$ as a $\mathbb{Z}/n\mathbb{Z}$-module, and I didn't know how to do that, so asked about it in an earlier question (What is a projective resolution of $\mathbb{Z}/m\mathbb{Z}$ as a $\mathbb{Z}/n\mathbb{Z}$-module?).
However, the answer seems to depend on whether $m$ has repeated prime factors of $n$, so this doesn't seem like the way to compute these Ext and Tor groups. Is there a better way to do this?
I'll write $Z_n$ for $\Bbb Z/n\Bbb Z$ to reduce typing. Then a projective resolution of $Z_m$ as a $Z_n$-module goes like this $$\cdots\to Z_n\xrightarrow{\times n/m} Z_n\xrightarrow{\times m} Z_n\xrightarrow{\times n/m} Z_n\to Z_m\to0.$$ Dropping the final $Z_m$ and tensoring with $Z_m$, the Tor groups are the homology groups of $$\cdots\to Z_m\xrightarrow{\times n/m} Z_m\xrightarrow{\times m} Z_m\xrightarrow{\times n/m} Z_m\to0.$$ Then $\times m$ maps are 0, so the even Tor groups are all equal the cokernel of the multiplication of $n/m$ on $Z_m$ and the odd Tor groups equal the kernel of this map. I don't have the energy to compute these further or look at the Ext groups...