I'm skimming through Silverman's text to recall some theory of elliptic curves that I've learned in undergrad. In practice however, I'm having trouble actually computing the cohomology groups. For instance, I know intuitively that $H^1(\mathbb{F}_q,E)$ will vanish, but how does one show this? Analogously, how does one compute $H^1(\mathbb{R},E)$?
I know that for a field $k$ and $n\in\mathbb{Z}$, the corresponding map $n:E(\bar k)\to E(\bar k)$ is surjective. This gives an exact sequence \begin{equation} 0\to E(\bar k)[n]\to E(\bar k)\to^n E(\bar k)\to 0, \end{equation} which further gives the exact sequence \begin{equation} 0\to E(k)[n]\to E(k)\to^n E(k)\to H^1(k,E(\bar k)[n])\\ \hspace{5em}\to H^1(k,E(\bar k))\to^n H^1(k,E(\bar k)). \end{equation} Hence from the long exact sequence we have \begin{equation} 0\to E(k)/nE(k)\to H^1(k,E(\bar k)[n])\to H^1(k,E(\bar k))[n]\to 0. \end{equation} How can one use this to compute cohomology over a finite field or real numbers?
Both calculations you ask about are done in this lecture note of mine on WC-groups: see pages 5-7. Mariano's remark is used in the former computation; it is also what I have in mind in "Exercise 6", although if you keep reading you'll see the computation redone several more times in fancier ways.
(A proof of Theorem 5 of the above note appears in $\S 3$ of this eternally not quite finished paper of mine.)
$\newcommand{\F}{\mathbb{F}}$ Added: Another slick proof that $H^1(\F_q,E) = 0$ is to note that a nonzero element would yield a genus one curve $C_{/\mathbb{F}_q}$ without a $\F_q$-rational point. But Weil proved that for any curve $C_{/\mathbb{F}_q}$ of genus $g$, $|\# C(\F_q) - (q+1)| \leq 2g \sqrt{q}$. When $g = 1$ this implies $\# C(\F_q) > 0$(!!).