Find a chain complex $C_*$ such that $H_i(C_*)=0$ for all $i$, but $H_i(C_*\otimes \mathbb Z_3)\neq 0$ for some $i$.
Recall that if $C_*$ is a chain complex with differential $\partial$, $C_*\otimes A$ is a chain complex such that $(C_*\otimes A)_n=C_n\otimes A$ and the differential is given $\partial_n(c\otimes a)= \partial_n(c)\otimes a$.
The first part of this question asked to find an example of a space $X$ such that: $H_i(X,\mathbb Z)$ was non trivial for some $i>0$ and $H_i(X,\mathbb Z_3)=0$ for all $i>0$. I found that $X=\mathbb RP^4$ satisfies that condition.
It seems like there should be an easy way to this part of the question givent he first part but it almost seems backwards.
My idea so far was to let our chain be something like $\mathbb Z \rightarrow \mathbb Z \rightarrow \mathbb Z $ where the first map is the identity and the second map is multiplication by 3. Since these are isomorphism we should get trivial homology?
Then when we tensor with $\mathbb Z_3$ we get the identity and then a zero map, which gives non trival homology. But I'm not sure how to make $H_i(C_*)=0$ for all $i$.
$\dots \rightarrow 0 \rightarrow \mathbb{Z} \xrightarrow{3} \mathbb{Z} \rightarrow \mathbb{Z}/3 \rightarrow 0 \rightarrow \dots$ is exact, and so this has 0 homology. As you said, tensoring by $\mathbb{Z}/3$ will cause the new complex to not be exact, and so it has nontrivial homology. It is worth noting that no space's singular, simplicial, or cellular chain complex will give a counterexample by the universal coefficient theorem.