Let $\left\{a_n\right\}$ be a decreasing sequence of real numbers such that $n\cdot a_n\to 0$. Then the series $\sum_{n\geqslant 2}a_n\sin (nx)$ is uniformly convergent on $\mathbb R$.
Thanks to Abel transform, we can show that the convergence is uniform on $[\delta,2\pi-\delta]$ for all $\delta>0$. Since the functions are odd, we only have to
prove the uniform convergence on $[0,\delta]$. Put $M_n:=\sup_{k\geqslant n}ka_k$, and
$R_n(x)=\sum_{k=n}^{\infty}a_k\sin (kx)$. Fix $x\neq 0$ and $N$ such that $\frac 1N\leqslant x<\frac 1{N-1}$. Put for $n\lt N$:
$$A_n(x)=\sum_{k=n}^{N-1}a_k\sin kx\mbox{ and }B_n(x):=\sum_{k=n}^{+\infty}a_k\sin (kx),$$
and for $n\geq N$, $A_n(x):=0$.
Since $|\sin t|\leqslant t$ for $t\geq 0$ we have
$$|A_n(x)|\leqslant \sum_{k=n}^{N-1}a_kkx\leqslant M_nx(N-n)\leqslant \frac{N-n}{N-1}M_n,$$
so $|A_n(x)|\leqslant M_n$.
If $N> n$, we have after writing $D_k=\sum_{j=0}^k\sin jx$, $|D_k(x)|\leqslant \frac cx$ on $(0,\delta]$ for some constant $c$. Indeed, we have
$|D_k(x)|\leqslant \frac 1{\sqrt{2(1-\cos x)}}$ and $\cos x=1-\frac{x^2}2(1+\xi)$ where
$|\xi|\leqslant \frac 12$ so $2(1-\cos x)\geqslant \frac{x^2}2$ and $|D_k(x)\leqslant \frac{\sqrt 2}x$. Therefore
$$|B_n(x)|\leqslant \frac{\sqrt 2}x\sum_{k=N}^{+\infty}(a_k-a_{k+1})+a_N\frac{\sqrt 2}x=\frac{2\sqrt 2}xa_N\leqslant 2\sqrt 2 Na_N\leqslant 2\sqrt 2M_n.$$
We get the same bound if $N\leqslant n$. Finally $|R_n(x)|\leqslant (2\sqrt 2+1)M_n$ for all $0\leqslant x\leqslant \delta$, so the convergence is uniform on $\mathbb R$.
Added latter: it's an example of a Fourier series which is uniformly convergent on the real line, but not absolutely convergent at any point of $(0,2\pi)$. Indeed take $x\in (0,2\pi)$. Since $|\sin(nx)|\geqslant \sin^2(nx)$, we would have the convergence of $\sum_{n\geqslant 2}\frac{\sin^2(nx)}{n\log n}$. We have $\sin ^2(nx)= \frac 1{-4}(e^{inx}-e^{-inx})^2=-\frac 14 (e^{2inx}+e^{-2inx}-2)=\frac 12-\frac 12\cos (2nx)$ and an Abel transform shows that the series $\sum_{n\geqslant 2}\frac{\cos(2nx)}{n\log n}$ is convergent. So the series $\sum_{n\geqslant 2}\frac 1{n\log n}$ would be convergent, which is not the case as the integral test shows.
Best Answer
For part (b), consider a sequence $x_n = \frac{\pi}{4n} + 2n\pi \in E_2$. For $n < k \leqslant 2n$ we have
$$\frac{1}{\sqrt{2}} = \sin \left(\frac{\pi}{4}+ 2n^2 \pi \right) = \sin nx_n < \sin k x_n < \sin 2nx_n = \sin \left(\frac{\pi}{2}+ 2n^2 \pi \right) = 1$$
Thus,
$$ \left|\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} \right| =\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} > n \cdot \frac{x_n}{2n + x_n}\cdot \frac{1}{\sqrt{2}} = \frac{2n^2\pi + \frac{\pi}{4}}{\sqrt{2}(2n + 2n\pi + \frac{\pi}{4n})} $$
Since, the RHS tends to $+\infty$ as $n \to \infty$, the series cannot be uniformly convergent on $E_2$, by violation of the uniform Cauchy criterion.
Edit (9/16/2022):
The argument above is flawed because with $x_n = \frac{\pi}{4n} + 2n\pi$, we should have $\frac{\pi}{4} +2n^2\pi < kx_n \leqslant \frac{\pi}{2} + 2n^2\pi + 2n^2\pi$ and we don't have $\sin kx_n > 0$ for all $n< k\leqslant 2n$.
It should be amended as follows.
Consider a sequence $x_n = \frac{\pi}{4n} + 2\pi \in E_2$. For $n < k \leqslant 2n$ we have $\frac{\pi}{4} < \frac{\pi k}{4n}\leqslant \frac{\pi}{2}$ and
$$\frac{1}{\sqrt{2}} = \sin \left(\frac{\pi}{4}\right) < \sin \frac{\pi k}{4n} =\sin \left(\frac{\pi k}{4n}+ 2k\pi\right) = \sin kx_n < \sin \left(\frac{\pi}{2} \right) = 1$$
Thus,
$$ \left|\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} \right| =\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} > n \cdot \frac{x_n}{2n + x_n}\cdot \frac{1}{\sqrt{2}} = \frac{2n\pi + \frac{\pi}{4}}{\sqrt{2}(2n + 2\pi + \frac{\pi}{4n})} $$
Since the RHS tends to $\frac{\pi}{\sqrt{2}}\neq 0$ as $n \to \infty$, the series cannot be uniformly convergent ...