If $G$ is a PRG, is $F(s)=G(S)||G(\bar{s})$ necessarily a PRG?
$F(s)=G(s)\mathbin\|G(\bar{s})$ is not necessarily a PRG. We'll construct a counterexample.

Assume $H(t)$ is a PRG.

Now define $G(t\mathbin\|0)=H(t)$ and $G(t\mathbin\|1)=H(\bar t)$. The $G(s)$ thus defined is a PRG with input width $|t|+1$. Yet $F(s)=G(s)\mathbin\|G(\bar s)$ is not a PRG, because its two halves are always equal.