First, denote the integral below as $I$$$I=\int\limits_0^1dx\space\frac {x(1-x)}{\sin\pi x}$$and through integration by parts on $u=x-x^2$, then we have
$$\begin{align*}I & =-\frac 1{\pi}(x-x^2)\log\cot\left(\frac {\pi x}2\right)\,\Biggr\rvert_0^1+\frac 1{\pi}\int\limits_0^1dx\, (1-2x)\log\cot\left(\frac {\pi x}2\right)\\ & =\frac 1{\pi}\int\limits_0^1dx\,\log\cot\left(\frac {\pi x}2\right)-\frac 2{\pi}\int\limits_0^1dx\, x\log\cot\left(\frac {\pi x}2\right)\\ & =-\frac 8{\pi^3}\int\limits_0^{\pi/2}dx\, x\log\cot x\tag1\end{align*}$$
where equation ($1$) comes from making the substitution $x\mapsto\frac {\pi x}2$. The latter integral can be evaluated by splitting up the natural log into two separate integrals and using the Fourier series for $\log\sin x$ and $\log\cos x$, which I have included below
$$\begin{align*}\log\cos x & =\sum\limits_{k\geq1}(-1)^{k-1}\frac {\cos2kx}{k}-\log 2\tag2\\\log\sin x & =-\sum\limits_{k\geq1}\frac {\cos 2kx}k-\log 2\tag3\end{align*}$$
Expanding ($1$) gives
$$I=-\frac 8{\pi^3}\underbrace{\int\limits_0^{\pi/2}dx\, x\log\cos x}_{I_1}+\frac 8{\pi^3}\underbrace{\int\limits_0^{\pi/2}dx\, x\log\sin x}_{I_2}\tag4$$
Call the first and second integrals $I_1$ and $I_2$ respectively. Using ($2$) and ($3$) gives the following two identities
$$\begin{align*}I_1 & =\int\limits_0^{\pi/2}dx\,\left(\sum\limits_{k\geq1}\frac {(-1)^{k-1}\cos 2kx}k-x\log 2\right)\\ & =\sum\limits_{k\geq1}\frac {(-1)^{k-1}}k\left[\frac {\pi}{4k^2}\sin\pi k+\frac 1{4k^3}\cos\pi k-\frac 1{4k^2}\right]-\frac {\pi^2}8\log2\\ & =\frac 14\sum\limits_{k\geq1}\frac {(-1)^{k-1}}{k^3}\cos\pi k-\frac 14\sum\limits_{k\geq1}\frac {(-1)^{k-1}}{k^3}-\frac {\pi^2}8\log 2\\ & \color{blue}{=-\frac 14\zeta(3)-\frac 3{16}\zeta(3)-\frac {\pi^2}8\log 2}\tag5\end{align*}$$
As a side note, the infinite sum with $\sin\pi k$ vanishes because $\sin\pi k=0$ for $k\in\mathbb{Z}$. In a similar manner, $I_2$ can be integrated as follows
$$\begin{align*}I_2 & =-\int\limits_0^{\pi/2}dx\,\left(\sum\limits_{k\geq1}\frac {\cos 2kx}k+x\log 2\right)\\ & =-\sum\limits_{k\geq1}\frac 1k\left[\frac {\pi}{4k}\sin\pi k+\frac 1{4k^2}\cos\pi k-\frac 1{4k^2}\right]-\frac {\pi^2}8\log 2\\ & =-\frac 14\sum\limits_{k\geq1}\frac {\cos\pi k}{k^3}+\frac 14\sum\limits_{k\geq1}\frac 1{k^3}-\frac {\pi^2}8\log 2\\ & \color{red}{=\frac 14\zeta(3)+\frac 3{16}\zeta(3)-\frac {\pi^2}8\log 2}\tag6\end{align*}$$
Substituting the results for ($5$) and ($6$) into ($4$) leaves us with
$$\begin{align*}I & =-\frac 8{\pi^3}\left[\color{blue}{-\frac 14\zeta(3)-\frac 3{16}\zeta(3)}\color{red}{-\frac 14\zeta(3)-\frac 3{16}\zeta(3)}\right]\\ & =\frac 7{\pi^3}\zeta(3)\end{align*}$$
Multiply by $-1$ to get the integral under question
$$\int\limits_0^1dx\space\frac {x^2-x}{\sin\pi x}\color{brown}{=-\frac 7{\pi^3}\zeta(3)}$$
As shown by @eyeballfrog,
$$\int_0^{\infty} \frac{e^{\cos(x)}\sin(\sin(x))}{x} dx =\int_0^\pi \frac{\sin[\sin(u)]}{\sin(u)}\sin^2\left(\frac{u}{2}\right)e^{-\cos(u)}du$$
Further manipulating:
$$\begin{align}
\int_0^\pi \frac{\sin[\sin(u)]}{\sin(u)}\sin^2\left(\frac{u}{2}\right)e^{-\cos(u)}du
&=\int_0^\pi \frac{\sin[\sin(u)]}{\sin(u)}\left(\frac{1-\cos u}{2}\right)e^{-\cos(u)}du \\
&=\frac12\int_{-\pi}^\pi \frac{\sin[\sin(u)]}{\sin(u)}\left(\frac{1-\cos u}{2}\right)e^{-\cos(u)}du \\
&=\frac14\Im\int_{-\pi}^\pi \frac{1-\cos u}{\sin(u)}e^{-\cos(u)+i\sin u}du \\
&=\frac14\Im\int_{-\pi}^\pi \frac{1-\cos u}{\sin(u)}\exp(-e^{-iu})du \\
&=-\frac14\Im\int_{-\pi}^\pi \frac{1-\cos u}{\sin(u)}\exp(-e^{iu})du \\
\end{align}
$$
Let $z=e^{iu}$, then
$$\begin{align}
\int_0^\pi \frac{\sin[\sin(u)]}{\sin(u)}\sin^2\left(\frac{u}{2}\right)e^{-\cos(u)}du
&=-\frac14\Im\int_{-\pi}^\pi \frac{2-2\cos u}{2\sin(u)}\exp(-e^{iu})du \\
&=-\frac14\Im\oint_{|z|=1} \frac{2-z-z^{-1}}{(z-z^{-1})/i}\exp(-z)\frac{dz}{iz} \\
&=\frac14\Im\oint_{|z|=1} \frac{z^2-2z+1}{z^2-1}\frac{e^{-z}}z dz\\
&=\frac14\Im\underbrace{\oint_{|z|=1} \frac{z-1}{z+1}\frac{e^{-z}}z dz}_{I}\\
\end{align}
$$
Note that the contour integral has to be understood in the Cauchy principal value sense, since a pole lies on the path of integration.
Consider the contour $C$, a unit circle with a semicircle indent to the right at $z=-1$.
By residue theorem, $$\oint_C \frac{z-1}{z+1}\frac{e^{-z}}z dz=2\pi i\operatorname*{Res}_{z=0}\frac{z-1}{z+1}\frac{e^{-z}}z$$
$$\implies I+\int_{\text{indent}}\frac{z-1}{z+1}\frac{e^{-z}}z dz=-2\pi i$$
Since the indent is a semicircle and goes clockwisely, it is not difficult to prove that
$$\int_{\text{indent}}\frac{z-1}{z+1}\frac{e^{-z}}z dz=-\frac12\cdot 2\pi i\operatorname*{Res}_{z=-1}\frac{z-1}{z+1}\frac{e^{-z}}z=-2\pi i\cdot e$$
Hence, $$I=2\pi i (e-1)$$
As a result,
$$\int_0^{\infty} \frac{e^{\cos(x)}\sin(\sin(x))}{x} dx=\frac\pi 2(e-1)$$
which has been confirmed numerically.
Best Answer
Both results follow from the well known Dirichlet integral $\int_0^{\infty}\frac{\sin x}{x}=\frac{\pi}{2}$
We first notice that $g(x)=\frac{1}{\log x}+\frac{1}{1-x}$ is continuous on $[0,1]$ (where we define by continuity $g(0)=1, g(1)=1/2$ since $1-x+\log x=-(1-x)^2/2+O((1-x)^3), \log x=-(1-x)(1+O(1-x))$ near $1^-$)
Hence by the Riemann-Lebesgue lemma:
$\lim_{m\to\infty}\int_0^1\sin(m\pi x)g(x)dx=0$ so the required limits are $-\lim_{n\to\infty}\int_0^1\frac{\sin(2n\pi x)}{1-x}dx$ and $-\lim_{n\to\infty}\int_0^1\frac{\sin((2n+1)\pi x)}{1-x}dx$ respectively once we show that those exists.
Changing variables $x \to 1-x$ and using $\sin(2n\pi (1-x))=-\sin(2n\pi x)$ the first integral becomes
$\int_0^1\frac{\sin(2n\pi x)}{x}dx=\int_0^{2n\pi}\frac{\sin y}{y}dy$ and the limit follows from the Dirichlet integral as noted.
Obviously $\sin((2n+1)\pi (1-x))=\sin((2n+1)\pi x)$ so the second integral preserves the minus sign and the limit is then $-\pi/2$