Here is an example to show that 1) does not hold (when $n=2$) : consider the measure $\mu=\sum n^{-2}\delta_{\alpha_{n}}$ where the sum runs over all rational numbers of $[-1,1]$, and denote by $U^{\mu}$ the associated logarithmic potential. Since $U^{\mu}$ is finite except on a polar set, the set
$$A_{n}=\{z\in[-1,1],~U^{\mu}(z)<n\},\qquad n\text{ large enough},$$
is of positive capacity. Moreover, the set
$$S_{n}=\{z\in\mathbb{C},~U^{\mu}(z)>n\}$$
is open and non-empty, it contains $\mathbb{Q}\cap[-1,1]$, and $A_{n}\subset \overline S_{n}\setminus S_{n}$ because each point of $A_{n}$ is a limit point of $\mathbb{Q}\cap[-1,1]$ and $U^{\mu}$ equals $\infty$ at those points. Finally, $S_{n}$ is thin at each point $\zeta$ of $A_{n}$ since
$$
\liminf_{z\to\zeta,~z\in S_{n}}U^{\mu}(z)\geq n>U^{\mu}(\zeta).
$$
This example shows that 2) is also false, because on $[-1,1]$, the points $z$ of
$\overline S_{n}\setminus A_{n}$ obviously satisfy $U^{\mu}(z)\geq n$, and outside of $[-1,1]$, the potential is continuous and thus the points of $\overline S_{n}$ also satisfy $U^{\mu}(z)\geq n$. Thus $\overline S_{n}\setminus A_{n}$ is thin at $A_{n}$.

TheoremLet $F$ be an $F_\sigma$ subset of $\mathbb{C}$. If $F$ is thin at $0$, then $E=\{ e^{i\theta}|\ r_{n}e^{i\theta}\in F, \text{for some sequence }r_n\rightarrow 0\}$ is a polar set. $\endgroup$ – Josiah Park Nov 5 '19 at 6:57