I am working on this problem, and here is what I figured out:
$\mathbb Z/ 6\mathbb Z \xrightarrow{\times 2} \mathbb Z/ 6\mathbb Z \xrightarrow{\times 3} \mathbb Z/ 6\mathbb Z \xrightarrow{\times2 } \mathbb Z/ 6\mathbb Z \rightarrow 0$ is a projective resolution of $\mathbb Z/2\mathbb Z$.
Then tensor it with $\mathbb Z/ 2 \mathbb Z$, we get the following sequence:
$\mathbb Z/ 2\mathbb Z \xrightarrow{0} \mathbb Z/ 2\mathbb Z \xrightarrow{id} \mathbb Z/ 2\mathbb Z \xrightarrow{0} \mathbb Z/ 2\mathbb Z \rightarrow 0$
So the left nth left derived functor would be zero if $n \neq 0$.
But there is a hint saying that we should use the formula $\mathbb Z/m \mathbb Z \otimes \mathbb Z / n \mathbb Z = \mathbb Z / gcd(m,n) \mathbb Z$ to get a closed formula which confuses me since I did not use it at all.