Here's another counterexample for Problem 1 not using the fancy (non-explicit) construction of Shelah.

Fix an odd prime $p$. Let $A=\mathbf{F}_p((t))$ be the ring of Laurent series over the field $\mathbf{F}_p$ on $p$ elements; we only view it as an abstract group, and let $A_0$ be the set of series $\sum a_nt^n$ in $A$ with $a_0=0$. Let $B$ be the non-unital subring $t^{-1}\mathbf{F}_p[t^{-1}]$. Define a symplectic form on $A_0$ by $\langle t^n,t^m\rangle=0$ if $n+m=0$ and $\langle t^n,t^{-n}\rangle=1$ for all $n\ge 0$, by extending it by "formal" linearity (namely
$$ \langle\sum a_n t^n,\sum b_nt^m\rangle=\sum_{n>0}a_nb_{-n}-a_{-n}b_n\qquad\qquad ).$$
That it is non-degenerate is straightforward. Observe that $B$ is a maximal isotropic subspace.

Define a (nilpotent) uncountable group $G$ as the set of pairs $(x,t)$ with $x\in A_0$, $t\in\mathbf{F}_p$, and group law $(x,t)(x',t')=(x+x',t+t'+\langle x,x'\rangle)$. Then $B$ is a countable subgroup and every group properly containing $B$ contains $(0,1)\notin B$.