This is surely well known. Let $X$ and $Y$ be smooth, projective, connected complex varieties. Then $$H^2(X\times Y,Z/n)=H^2(X,Z/n)\oplus H^2(Y,Z/n) \oplus (H^1(X,Z/n)\otimes H^1(Y,Z/n))$$ for any $n>1$. I think I can prove this using a counting argument and the K\"unneth formula with coefficients in $Z$, but it seems a ridiculous way of doing it. Is there a more natural way or a good reference?
The same question for the \'etale cohomology of varieties over a separably closed field of characteristic not dividing $n$. (Same comment)



The identity map of $H^2(X,\mathbb{Z}/n)\oplus H^2(Y,\mathbb{Z}/n)$ can be decomposed as $$H^2(X,\mathbb{Z}/n)\oplus H^2(Y,\mathbb{Z}/n)\to H^2(X\times Y,\mathbb{Z}/n)\to H^2(X,\mathbb{Z}/n)\oplus H^2(Y,\mathbb{Z}/n)$$ where the first arrow is the sum of the pullbacks and the second arrow is the sum of the maps induced by inclusions of $X,Y$ into the product. So $H^2(X,\mathbb{Z}/n)\oplus H^2(Y,\mathbb{Z}/n)$ splits off. The remaining part can be identified with $H^1(X,H^1(Y,\mathbb{Z}/n))$ via the Leray spectral sequence. Now let me consider the topological case. In this case we have $$H^1(Z,A)=Hom(H_1(Y,A),A))$$ for any nonpathological space $Z$ and any ring $A$ since $H_0(Z,A)$ is free. So we get $$H^1(X,H^1(Y,\mathbb{Z}/n))\cong Hom(H_1(X,\mathbb{Z}/n)\otimes H_1(Y,\mathbb{Z}/n),\mathbb{Z}/n)$$ $$\cong Hom (H_1(X,\mathbb{Z}),\mathbb{Z}/n)\otimes Hom (H_1(Y,\mathbb{Z}/n),\mathbb{Z}/n)\cong H^1(X,\mathbb{Z}/n)\otimes H^1(Y,\mathbb{Z}/n).$$ 

