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 $$ \liminf_{z\to\zeta,~z\in S_{n}}U^{\mu}(z)\geq n>U^{\mu}(\zeta). $$

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 $A_{n}$.

1. By Theorem 4.2.4 of the reference below, the set $D$ is thin at $z\in\partial D$ if and only if $z$ is an irregular point of the complement $D^{c}$. Moreover, by Theorem 4.2.5, the set of irregular boundary points of a domain is polar.

2. Let $D$ be the union of the unit disk $\mathbb{D}$ and the segment $E=[2,3]$. Then, $\overline D\setminus E=\mathbb{D}$ is thin at each point of $E$ and $E$ is not polar.

T. Ransford, Potential theory in the complex plane, Cambridge, 1995.

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 $$ \liminf_{z\to\zeta,~z\in S_{n}}U^{\mu}(z)\geq n>U^{\mu}(\zeta). $$