(1) An extension of the finitely generated case for Q1: let $K_0$ be finitely generated over the prime field, and let $K=K_0((x_i)_{i\in I})$ be a purely transcendental extension of $K_0$. Then $K$ "satisfies Q1". Indeed, any abelian variety $A/K$ is defined over some intermediate $K_1:=K_0((x_i)_{i\in J})$, $J\subset I$ finite. Then $A(K_1)$ is finitely generated, but $A(K)=A(K_1)$ since $K/K_1$ is purely transcendental.
(2) Another "easy" case for Q2: if $K/\mathbb{F}_p$ is infinite algebraic, then for any $A$ the group $A(K)$ is torsion, but must be infinite by Weil's estimates, hence is not finitely generated.
(3) Some more general remarks on Q2: assume $K$ is a field and $A$ an abelian variety over $K$ such that $A(K)$ is finitely generated. Take any rational function $f$ on $A$. Then there is a subfield $K_0$ of $K$, finitely generated over the prime field, such that $f(A(K))\subset K_0$ (of course I exclude points where $f$ is undefined). Indeed, take for $K_0$ a field of definition for $A$, $f$, and generators of $A(K)$.
This proves that other examples for Q2 are:
(3a) all Henselian valued fields (already mentioned by Pete, but there is no restriction on the rank here, except the valuation must be nontrivial). Indeed, take $f$ defined and smooth at the origin, and (say) sending the origin to $0$. By the implicit function theorem, $f(A(K)$ contains a neighborhood $V$ of $0$ for the valuation topology; the subfield generated by $V$ is $K$.
(3b) Pseudo-algebraically closed (PAC) fields: indeed, by Bertini, there is an $f$ whose general fiber is smooth and geometrically connected, hence has rational points. (This works directly if $\dim A\geq2$; if $\dim A=1$, consider $A\times A$). This includes example (2) above.