This is much easier than it looks. The point is that any reductive group $G$ is isogenous to the product of its radical, which is its centre $Z(G)$, and its commutator subgroup, which is a semisimple group. Since Hom into $\mathbb{R}$ will kill all torsion, it is sufficient to prove the statement when $G$ is either semisimple, or a torus.
The semisimple case is trivial because both spaces are obviously 0. The torus case is slightly more delicate, but it amounts to the statement that the natural map from the maximal $\operatorname{Gal}(\overline{F} / F)$-invariant subspace of $X(T)$ to the maximal invariant quotient is an isomorphism (which is obvious, since the action of $\operatorname{Gal}(\overline{F} / F)$ factors through a finite group, and real representations of finite groups are completely reducible).
(PS In your second example something is a little fishy since the torus $A_0$ is neither $F$-split nor contained in the radical of $G$. But it is true that the maximal $F$-split torus in $Z(G)$ is trivial.)
