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)
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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 non-pathological space $Z$ and any ring $A$ since $H_0(Z,A)$ is free. So we get
$$H^1(X,H^1(Y,\mathbb{Z}/n))\cong Hom(H_1(X,\mathbb{Z}/n)\otimes H_1(Y,\mathbb{Z}/n),\mathbb{Z}/n)$$
$$\cong Hom (H_1(X,\mathbb{Z}),\mathbb{Z}/n)\otimes Hom (H_1(Y,\mathbb{Z}/n),\mathbb{Z}/n)\cong H^1(X,\mathbb{Z}/n)\otimes H^1(Y,\mathbb{Z}/n).$$
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$\begingroup$ I wish there was an answer that does not use homology. Then it would be easier to adopt it to the etale setting. $\endgroup$ Commented Oct 21, 2011 at 15:07
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$\begingroup$ Alexei -- I think there was an Asterisque volume that came out some time in the 70's or 80's called "Homologie \'etale" by Deligne, Verdier and other authors. As far as can I remember, they define \'etale homology with coefficients in a constructible sheaf as the compactly supported cohomology of the Verdier dual of the sheaf, with the opposite grading. It turns out this homology has many of the properties of the topological homology, but I do not remember the details. $\endgroup$– algoriCommented Oct 21, 2011 at 16:10
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$\begingroup$ By the way, one thing is not completely clear to me: one probably needs to prove the commutativity of a certain diagram, namely, that the injective map from $H^1(X,H^1(Y,Z/n))$ to $H^2(X\times Y,Z/n)$ is the same as the cup-product map from $H^1(X,Z/n)\otimes H^1(Y,Z/n)$. You show that the two groups are isomorphic, but maybe an abstract isomorphism is not enough. $\endgroup$ Commented Oct 21, 2011 at 19:45