The derived group $G'$ of $G$ always works for your second question (i.e., $G'(F)Z(F)$ is closed and cocompact in $G(F)$). Indeed, by using local class field theory and Kneser-Bruhat-Tits we know that ${\rm{H}}^1(F,H)$ is finite for any connected reductive $F$-group $H$, so ${\rm{H}}^1(F,G')$ is finite. If $X \rightarrow Y$ is any smooth map of smooth $F$-schemes then $X(F) \rightarrow Y(F)$ is an $F$-analytic submersion, so it has open image. Thus, the image of $G(F)$ in the commutative $(G/G')(F)$ is open, and it has finite index since ${\rm{H}}^1(F,G')$ is finite. In other words, $G(F)/G'(F)$ is a finite-index open subgroup of $(G/G')(F)$, so $G(F)/G'(F)Z(F)$ is compact if and only if the map $Z(F) \rightarrow (G/G')(F)$ induced by the surjection of $F$-tori $Z \rightarrow G/G'$ has compact cokernel.

Thus, it suffices to show that for any surjective map $T' \rightarrow T$ between $F$-tori (even inseparable), the map $T'(F) \rightarrow T(F)$ has closed image with compact cokernel. The maximal compact subgroup of $T(F)$ is
$$T(F)^1 = \cap_{\chi\in {\rm{X}}_F(T)} \ker |\!|\chi|\!|_F$$
where $\chi$ varies through the $F$-rational cocharacters of $T$ and $|\!| \cdot |\!|_F$ is the normalized absolute value. In other words, $T(F)^1$ is the group of $t \in T(F)$ such that $\chi(t) \in O_F^{\times}$ for all such $\chi$.
It is harmless to pass to quotients by maximal compact subgroups, which is to say that it is equivalent to show that the map $T'(F)/T'(F)^1 \rightarrow T(F)/T(F)^1$ has closed image with finite index.

But $T(F)^1$ is always open in $T(F)$ with the discrete cokernel $T(F)/T(F)^1$ naturally isomorphic to the $F$-rational cocharacter group ${\rm{X}}_{\ast,F}(T)$, so we are reduced to proving that ${\rm{X}}_{\ast,F}(T') \rightarrow {\rm{X}}_{\ast,F}(T)$ has image with finite index. Since these cocharacter groups are finitely generated $\mathbf{Z}$-modules and any surjection of $F$-tori admits an $F$-rational section *in the isogeny category*, we're done.