Suppose $A$ is an abelian variety over a number field $K$ and call $M$ the maximal torsion free quotient of $A(\overline{K})$ equipped with its Galois action.



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> For the first Galois cohomology of $M$, do we have $H^1(\text{Gal}\_K, M) = 0$? Is $H^1(\text{Gal}\_K, M)$ at least finite?
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If $H^1(\text{Gal}\_K/U, M^U)$ is trivial for all open normal subgroups of $\text{Gal}\_K$, then so is $H^1(\text{Gal}\_K, M)$. If the vanishing is false, I'd be interested in a counterexample evev just for some $U$ and $H^1(\text{Gal}\_K/U, M^U)$.



Galois cohomology of abelian varieties
I think that all Galois cohomology groups $\mathrm{H}^i(\mathrm{Gal}\_K,M)$ vanish for $i>0$. As you observed, it suffices to prove that
$\mathrm{H}^i(\mathrm{Gal}\_K/U,M^U)$ vanishes for all open normal subgroups $U$ of $\mathrm{Gal}\_K$. Since $\mathrm{Gal}\_K/U$ is a finite
group, the Galois cohomology group $\mathrm{H}^i(\mathrm{Gal}\_K/U,M^U)$ is torsion (via restriction/corestriction with the trivial subgroup,
see Serre: Corps locaux, p. 138). In fact, all of its elements have order a divisor of $n=|\mathrm{Gal}\_K/U|$.
It suffices then to show that the multiplication-by-$n$ map $[n]$ on $M^U$ is bijective.


The map $[n]\colon M^U\rightarrow M^U$ is injective since $M$ is torsion free. In order to show that it is surjective choose $x\in M^U$ and
consider the short exact sequence defining $M$
$$
0\rightarrow A(\bar K)\_{\mathrm{tor}}\rightarrow A(\bar K)\overset{\pi}{\rightarrow} M\rightarrow 0,
$$
where $A(\bar K)\_{\mathrm{tor}}$ is the torsion subgroup of $A(\bar K)$. Since $x\in M$ and $A(\bar K)$ is a divisible group, there is an element
$y\in A(\bar K)$ such that $\pi([n]y)=x$, where $[n]$ also denotes the multiplication-by-$n$ morphism on the abelian group $A(\bar K)$. Let us
prove that $\pi(y)\in M^U$. Choose $g\in U$. One has
$$
\pi([n]gy)=\pi(g[n]y)=g\pi([n]y)=gx=x=\pi([n]y)
$$
since $x\in M^U$. It follows that $[n](gy-y)\in\ker(\pi)$. Hence $[n](gy-y)$ is a torsion element of $A(\bar K)$. Then $gy-y$ is in $A(\bar
K)\_{\mathrm tor}$ as well. This means that
$$
g\pi(y)-\pi(y)=\pi(gy-y)=0,
$$
i.e., $g\pi(y)=\pi(y)$.
Hence $\pi(y)\in M^U$. Since $[n]\pi(y)=\pi([n]y)=x$, the morphism $[n]\colon M^U\rightarrow M^U$ is surjective indeed.


