We know that there is a projective resolution $$\cdots \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \xrightarrow{\times 5} \mathbb{Z} \rightarrow \mathbb{Z}/5\mathbb{Z}$$
Taking this resolution with $Hom(\_, \mathbb{Z}/7\mathbb{Z})$ gives us $$0 \rightarrow Hom(Z/5Z, Z/7Z) \rightarrow Hom(Z, Z/7Z) \xrightarrow{\times 5} Hom(Z, Z/7Z) \rightarrow Hom(Z, Z/7Z) \rightarrow \cdots$$
However, this is the same thing as $$0 \rightarrow 0 \rightarrow Z/7Z \xrightarrow{\times 5} Z/7Z \rightarrow Z/7Z \cdots$$
Then $Ext_Z^i = 0$ for all $i \ge 0$, because the maps must alternate between $\times 5$ and $\times 7$.
To calculate $Tor_i^Z$, I tensored the original resolution by Z/7Z and simplified to get $$\cdots \rightarrow Z/7Z \rightarrow Z/7Z \xrightarrow{\times 5} Z/7Z \rightarrow 0$$ This also gave me $Tor_i^Z = 0$ for all $i \ge 0$.
I am not sure if I did everything correctly because having all Tor and Ext being 0 seems sketchy.
Your work is mostly ok, but there's one thing here you need to be careful with; if $P_*$ is your projective resolution, you should be taking the homology/cohomology of $P_*\otimes_{\mathbb{Z}}\mathbb{Z}/7\mathbb{Z}$ and $\operatorname{Hom}(P_*,\mathbb{Z}/7\mathbb{Z})$, without the augmentation map $\mathbb{Z}\to\mathbb{Z}/7\mathbb{Z}$ (see the Wiki page).
Here's what I mean, and I'll use a more general case to illustrate. We have the following projective reolution of $\mathbb{Z}/n\mathbb{Z}$: $$0\xrightarrow{\ \ \ }\mathbb{Z}\xrightarrow{\ n \ }\mathbb{Z}\xrightarrow{\ \pi \ }\mathbb{Z}/n\mathbb{Z}\xrightarrow{\ \ \ }0.$$ So, for $\mathbb{Z}/m\mathbb{Z}$, the group $\operatorname{Tor}_k^{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z})$ is given by the homology of the chain complex: $$0\xrightarrow{\ \ \ }\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}/m\mathbb{Z}\xrightarrow{\ id\otimes n \ }\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}/m\mathbb{Z}\xrightarrow{\ \ \ }0$$ and $\operatorname{Ext}_{\mathbb{Z}}^k(\mathbb{Z}/n\mathbb{Z},G)$ will be the cohomology of the cochain complex: $$0\xrightarrow{\ \ \ }\operatorname{Hom}(\mathbb{Z},\mathbb{Z}/m\mathbb{Z})\xrightarrow{\ -\circ n \ }\operatorname{Hom}(\mathbb{Z},\mathbb{Z}/m\mathbb{Z})\xrightarrow{\ \ \ }0.$$ Note that we applied the functors $-\otimes_{\mathbb{Z}}\mathbb{Z}/m\mathbb{Z}$ and $\operatorname{Hom}(-,\mathbb{Z}/m\mathbb{Z})$ to the complex: $$0\xrightarrow{\ \ \ }\mathbb{Z}\xrightarrow{\ n \ }\mathbb{Z}\xrightarrow{\ \ \ }0$$ with the augmentation map omitted. As you noted, the above two complexes are isomorphic to $$0\xrightarrow{\ \ \ }\mathbb{Z}/m\mathbb{Z}\xrightarrow{\ n \ }\mathbb{Z}/m\mathbb{Z}\xrightarrow{\ \ \ }0.$$ Thus, taking homology and cohomology gives: $$\operatorname{Tor}_k^{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z})\cong\begin{cases} \dfrac{\mathbb{Z}/m\mathbb{Z}}{n(\mathbb{Z}/m\mathbb{Z})},\ &k=0 \\[5pt] \operatorname{Ker}(n),\ &k=1 \\[5pt] 0,\ &k\geq 2 \end{cases}$$ and $$\operatorname{Ext}_{\mathbb{Z}}^k(\mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z})\cong\begin{cases} \operatorname{Ker}(n),\ &k=0 \\[5pt] \dfrac{\mathbb{Z}/m\mathbb{Z}}{n(\mathbb{Z}/m\mathbb{Z})},\ &k=1 \\[5pt] 0,\ &k\geq 2 \end{cases}.$$ Setting $n=5$ and $m=7$, the Ext and Tor groups are all zero as you suspected, since $\gcd(5,7)=1$.