First note that
$$
\left|\frac{1}{2-\sin(nx)}\right|\le1\tag{1}
$$
Next note that for $z=\tan{\frac{x}{2}}$
$$
\begin{align}
\int_0^{2\pi}\frac{\mathrm{d}x}{2-\sin(x)}
&=\int_{-\infty}^\infty\frac{\frac{2\;\mathrm{d}z}{1+z^2}}{2-\frac{2z}{1+z^2}}\\
&=\int_{-\infty}^\infty\frac{\mathrm{d}z}{1-z+z^2}\\
&=\int_{-\infty}^\infty\frac{\mathrm{d}z}{\frac{3}{4}+(z-\frac{1}{2})^2}\\
&=\frac{2\pi}{\sqrt{3}}\tag{2}
\end{align}
$$
Thus, the average of $\frac{1}{2-\sin(x)}$ over one of its periods is $\frac{1}{\sqrt{3}}$.
Approximate $F$ by a finite set of intervals $\{I_k\}$ so that $|\cup I_k\;\Delta\;F|<\epsilon$. Suppose $I_k=[a,b]$, then
$$
\begin{align}
\lim_{n\to\infty}\int_{I_k}\frac{\mathrm{d}x}{2-\sin(nx)}
&=\lim_{n\to\infty}\int_a^b\frac{\mathrm{d}x}{2-\sin(nx)}\\
&=\lim_{n\to\infty}\frac{1}{n}\int_{na}^{nb}\frac{\mathrm{d}x}{2-\sin(x)}\\
&=\lim_{n\to\infty}\frac{1}{n}\left(\int_{na}^{na+k2\pi}\frac{\mathrm{d}x}{2-\sin(x)}+\int_{na+k2\pi}^{nb}\frac{\mathrm{d}x}{2-\sin(x)}\right)\\
&=\lim_{n\to\infty}\frac{1}{n}\left(k\frac{2\pi}{\sqrt{3}}+J_n\right)\\
&=\frac{b-a}{\sqrt{3}}\\
&=\frac{|I_k|}{\sqrt{3}}\tag{3}
\end{align}
$$
In $(3)$, $k=\lfloor\frac{n(b-a)}{2\pi}\rfloor$ so that $|nb-(na+k2\pi)|<2\pi$ and therefore by $(1)$, we get $|J_n|<2\pi$. Thus, $\lim\limits_{n\to\infty}\frac{k}{n}=\frac{b-a}{2\pi}$ and $\lim\limits_{n\to\infty}\frac{J_n}{n}=0$.
Thus, again using $(1)$,
$$
\begin{align}
\lim_{n\to\infty}\left|\int_F\frac{\mathrm{d}x}{2-\sin(nx)}-\int_{\cup I_k}\frac{\mathrm{d}x}{2-\sin(nx)}\right|
&\le\lim_{n\to\infty}\int_{\cup I_k\;\Delta\;F}\frac{\mathrm{d}x}{|2-\sin(nx)|}\\
&\le\epsilon\tag{4}
\end{align}
$$
Summarizing,
$$
\begin{align}
\lim_{n\to\infty}\left|\int_F\frac{\mathrm{d}x}{2-\sin(nx)}-\frac{|F|}{\sqrt{3}}\right|
&\le\lim_{n\to\infty}\left|\int_F\frac{\mathrm{d}x}{2-\sin(nx)}-\int_{\cup I_k}\frac{\mathrm{d}x}{2-\sin(nx)}\right|\\
&+\lim_{n\to\infty}\left|\int_{\cup I_k}\frac{\mathrm{d}x}{2-\sin(nx)}-\frac{|\cup I_k|}{\sqrt{3}}\right|\\
&+\lim_{n\to\infty}\left|\frac{|\cup I_k|}{\sqrt{3}}-\frac{|F|}{\sqrt{3}}\right|\\
&\le\epsilon+0+\frac{\epsilon}{\sqrt{3}}\tag{5}
\end{align}
$$
therefore,
$$
\lim_{n\to\infty}\int_F\frac{\mathrm{d}x}{2-\sin(nx)}=\frac{|F|}{\sqrt{3}}\tag{6}
$$