Let $L/\mathbb{Q}$ be a Galois extension of degree $p$ and $E$ be an elliptic curve defined over $\mathbb{Q}$. Let $p$ be a fixed prime (of good ordinary reduction if required). We use $L_\infty, \mathbb{Q}_\infty$ as the notation for the cyclotomic $\mathbb{Z}_p$-extension. Let $w|p$ be primes in $L$. We have $L_{\infty,w}/ \mathbb{Q}_{\infty,p}$, a Galois extension and say the Galois group is denoted by $G$.
How does one one compute the Galois cohomology $H^i(G, \oplus_{w\mid p}H^1(L_{\infty,w},E_{p^\infty}))$ for $i=1,2$.
Computing $H^i(G, \oplus_{w\mid p}H^1(L_{\infty,w},E)_{p^\infty})$ for $i=1,2$ can be done relatively easily.
When $w$ splits in $L_\infty$, one can say $$ \bigoplus_{w\mid p}H^1(L_{\infty,w},E)_{p^\infty}\simeq H^1(K_{\infty,v},E)_{p^\infty} \otimes_{\mathbb{Z}_p} \mathbb{Z}_p[G] $$ and the right hand side is cohomologically trivial. We thus only care about when $w$ doesn't split in $L_\infty$. One can first of all show using local Tate duality (followed by Hochschild Serre) that for $i=1,2$ $$ H^i(G, \oplus_{w\mid v}H^1(L_{\infty,w},E)_{p^\infty})\simeq H^i(G, E(L_{\infty,w})) $$
My understanding is that using a result of Lang and a deep result of Coates-Greenberg one can say $H^i(G, E(L_{\infty,w}))=0$ for $i=1,2$. Consider the following short exact sequence $$ 0\rightarrow \mathcal{F}(\mathfrak{m}_w) \rightarrow E(L_{\infty,w}) \rightarrow \tilde{E}(\ell_w) \rightarrow 0. $$ Here $\mathcal{F}(\mathfrak{m}_w)$ is the corresponding formal group and $\ell_w$ the residue field. $\tilde{E}$ is reduction modulo $\mathfrak{m}_w$. The result of Lang says $H^i(G, \tilde{E}(\ell_w))=0$ for $i=1,2$ and that of Coates-Greenberg says $H^i(G,\mathcal{F}(\mathfrak{m}_w))=0$. This implies $H^i(G, E(L_{\infty,w}))=0$
cross-posted: https://math.stackexchange.com/questions/3082154/certain-galois-cohomology-computation