I was reading about algebraic groups and I ran into a proof that I don't understand. The setup is as follows: $k$ is a separably closed field, and $G$ is a simply connected semi-simple group over $k$. The claim is that $Pic(G)=\{1\}$, and the argument provided is:
Because $G$ is split reductive it is rational (I am guessing that rational means $Frac(O_G)=Frac(k[x_1,\dots, x_n])$ for some $n$). Because $G$ is rational then $Pic(G)=Ext^1(G,\mathbb{G}_m)$. And because $G$ is simply connected $Ext^1(G,\mathbb{G}_m)=\{1\}$.
The claims that I don't understand are:
What does being rational have to do with $Ext^1(G,\mathbb{G}_m)=Pic(G)$.
Why does being simply connected imply that $Ext^1(G,\mathbb{G}_m)=\{1\}$.