Problems 1 and 2 both have affirmative answers (implying that the finitary ballean of any uncountable group is normal).

Two cases are possible:

I. There exists a countable subgroup $A\subset G$ and an uncountable subset $B\subset G$ such that $bAb^{-1}\cap A$ is infinite for all $b\in B$. Replacing $B$ by a smaller uncountable set, we can assume that the family $(bA)_{b\in B}$ is disjoint. The latter condition can be used to show that the sets $A,B$ satisfy the condition 1 of the Problem.

We claim that for any function $\Phi:G\to [G]^{<\omega}$ there are elements $x,y\in G$ and $a\in A\setminus\Phi(x)$ and $b\in B\setminus \Phi(y)$ such that $xa=yb$.
Since $B$ is uncountable, there exists an element $b\in B\setminus\bigcup_{a\in H}\Phi(a)$. The set $bAb^{-1}\cap A$ is infinite and hence contains some element $a\notin\Phi(b)$. Put $x=b$ and $y=bab^{-1}\in A$. Observe that $xa=ba=yb$, $a\notin\Phi(b)=\Phi(x)$ and $b\notin\Phi(y)$.

II. There exists a countable infinite subgroup $A\subset G$ and an uncountable set $B'\subset G$ such that $bAb^{-1}\cap A$ is finite for every $b\in B'$. Replacing $B'$ by a smaller uncountable subset, we can aditionally assume that the family $(AbA)_{b\in B'}$ is disjoint.

We claim that the sets $A$ and $B=\{aba^{-1}:a\in A,\;b\in B'\}$ satisfy the conditions 1 and 2. The condition 1 will follow as soon as we check that for every $x\in G$ the intersection $xA\cap B$ is finite. Assuming that this intersection is not empty, we can find elemente $b\in B'$ and $a,\alpha\in A$ such that $x\alpha=aba^{-1}$. Taking into account that the family $(AvA)_{v\in B}$ is disjoint, we conclude that $xA\cap B=abA\cap B\subset abA\cap b^A$ where $b^A=\{gbg:g\in A\}$.

Given any element $z\in abA\cap b^A$, we can find elements $\alpha,g\in A$ with $ab\alpha=z=gbg^{-1}$ and conclude that $a^{-1}g=b\alpha gb^{-1}\in A\cap bAb^{-1}$. So, the element $g$ belongs to the finite set $F=a(A\cap gAg^{-1})$ and then $z=gbg^{-1}\in b^F:=\{fbf^{-1}:f\in F\}$. Therefore, $xA\cap B=abA\cap b^A$ is contained in the finite set $b^F$, which means that the sets $A,B$ satisfy the condition 1.

Now given any function $\Phi:G\to [G]^{<\omega}$, we shall find elements $x,y\in G$ and $a\in A\setminus\Phi(x)$ and $b\in B\setminus \Phi(y)$ such that $xa=yb$.

Using the uncountability of the set $B'$, choose an element $u\in B'\setminus \bigcup_{a\in A}a\Phi(a)a^{-1}$.
Find $a\in A\setminus \Phi(u)$. Put $y=a\in A$, $x=u$ and $b=a^{-1}ua\in B$. It follows that $xa=ua=aa^{-1}ua=yb$.
Also $a\notin\Phi(u)=\Phi(x)$ and $b=a^{-1}ua\notin \Phi(a)$ (as $u\notin a\Phi(a)a^{-1}$).

theJonsson group constructed by Shelah", as Shelah makes high use of the the axiom of choice and those various choices most likely produce many non-isomorphic Jonsson groups. $\endgroup$ – YCor Oct 18 '18 at 12:13