Let $p$ be a prime number and $\mathbb{Q}_p$ the field of $p$-adic numbers. Let $G_p$ be the absolute Galois group of $\mathbb{Q}_p$ and fix an absolutely irreducible representation $$ \rho : G_p \to GL_n(k)$$ where $k$ is a finite extension of $\mathbb{F}_p$.
If $p$ does not divide $n$, we can compute quite easily the group $H^2(G_p, Ad^0\rho(1))$ where $Ad^0\rho$ is the representation whose underlying space is the set of matrices in $M_n(k)$ of trace zero and with action give by conjugation by $\rho$. The $(1)$ means that we have twisted this action by the mod $p$ cyclotomic character. The computation is as follow :
We have a decomposition $Ad \rho = Ad^0 \rho \oplus k$ as representations of $G_p$ (because $p$ does not divide $n$) so that $Ad^0 \rho$ is self dual (because $Ad \rho$ and $k$ are). Hence, by Tate local duality theorem, we obtain that $H^2(G_p, Ad^0 \rho (1))$ is isomorphic to the dual of $H^0(Ad^0 \rho)$.
But $H^0(G_p, Ad^0 \rho)$ is zero because it is isomorphic to the set of matrices $T$ of trace zero such that $T \rho(g) = \rho(g) T$ for all $g \in G_p$, and by Schur lemma such matrices are necessarly scalar matrices. Again, because $p$ does not divide $n$, those scalar matrices are zero.
So we conclude that $H^2 (G_p, Ad^0 \rho(1)) = 0$ (I hope I did not make any mistakes...).
Now my question is as follow : what happens if $p$ does divide $n$ ?
In this case, $H^2 (Ad^0 \rho(1))$ is isomorphic to the dual of $H^0 (Ad \rho / k)$, but I don't know how to compute the later.