Let $A$ an abelian variety over a field $k$ and $A^{*}$ the dual abelian variety.
How can we relate the étale cohomology of $A$ with etale cohomology of $A^{*}$?
Let $A$ an abelian variety over a field $k$ and $A^{*}$ the dual abelian variety. How can we relate the étale cohomology of $A$ with etale cohomology of $A^{*}$? 


The Weil pairing induces a perfect pairing $T_\ell A \times T_\ell A^\vee \to \mathbf{Z}_\ell(1)$, and $T_\ell A = H^1_{et}(\bar{A},\mathbf{Z}_\ell)^\vee$ (here, $\bar{A}$ denotes $A \times_k \bar{k}$). Further, $H^i_{et}(\bar{A},\mathbf{Z}_\ell) = \wedge^iH^1_{et}(\bar{A},\mathbf{Z}_\ell)$. One also has $A^\vee = \mathcal{E}\mathrm{xt}_k^1(A,\mathbf{G}_m)$. 


Let me add something to what Timo Keller says rightly. What he said can be reformulated as $H^i_{et}(A,{\bf Z_\ell})$ and $H^i_{et}(A^*,{\bf Z_\ell})(1)$ are canonically dual of each other. This should be understand that there are dual not only as $\bf Z_\ell$modules, but also as $Gal(\bar k / k)$representation, and here the (1) takes its significance: it is a twist by the $\ell$adic cyclotomic character. Now, more can be said. Every abelian variety $A$ over $k$ as a polarization $p$ over $k$, which is in particular a symmetric isogeny $p: A \rightarrow A^\ast$. Now any isogeny induces isomorphism between the $V_\ell$'s (that is $T_\ell \otimes \bf Q_\ell$), hence taking again the wedge power, we get for all $i$ an isomorphism $$H^i_{et}(A,{\bf Z_\ell}) \otimes {\bf Q_\ell} \rightarrow H^i_{et}(A^*,\bf Z_\ell) \otimes \bf Q_\ell.$$ This isomorphism preserves other structures (the $Gal(\bar k/k)$action), but is not canonical, since it depends on the choice of $p$. But this is certainly a relation between the stale cohomology of $A$ and $A^\ast$ that is worth mentioning, and that is logically independent of the one given by Timo. 

