Note that we can write
$$\int_{-\infty}^\infty\frac{\sin(k)}{k}e^{ikx}\,dk=\frac{1}{2i}\int_{-\infty}^\infty\frac{e^{ik}-e^{-ik}}{k}e^{ikx}\,dk \tag1$$
Observe that for each of the principal value integrals
$$I_{\pm}(x)=\text{PV}\left(\int_{-\infty}^\infty\frac{e^{\pm ik}}{k}e^{ikx}\,dk\right)=\text{PV}\left(\int_{-\infty}^\infty\frac{e^{ ik(x\pm 1)}}{k}\,dk\right)$$
the integrand has a pole at $k=0$. To evaluate these integrals, we analyze the contour integrals
$$\begin{align}
\oint_{C}\frac{e^{iz(x\pm 1)}}{z}\,dz&=\int_{-R}^{-\epsilon} \frac{e^{ik(x\pm 1)}}{k}\,dk+\int_{\epsilon}^{R} \frac{e^{ik(x\pm 1)}}{k}\,dk\\\\
&+\int_{\text{sgn}(x\pm 1)\pi}^0 \frac{e^{i\epsilon e^{i\phi}}}{\epsilon e^{i\phi}}\,i\epsilon e^{i\phi}\,d\phi\\\\
&+\int_0^{\text{sgn}(x\pm 1)\pi}\frac{e^{i R e^{i\phi}}}{R e^{i\phi}}\,iR e^{i\phi}\,d\phi \tag 2
\end{align}$$
As $R\to \infty$, the last integral on the right-hand side of $(2)$ tends to $0$. As $\epsilon\to 0$, the third integral on the right-hand side of $(2)$ tends to $-\text{sgn}(x\pm 1)\pi$. Therefore, we have
$$\text{PV}\left(\int_{-\infty}^\infty\frac{e^{ ik(x+ 1)}-e^{ik(x-1)}}{k}\,dk\right)=\text{sgn}(x+1)\pi-\text{sgn}(x-1)\pi=\begin{cases}2\pi &,|x|<1\\\\0&,|x|>1\\\\\pi&,|x|=1\end{cases} \tag 3$$
Substituting $(3)$ into $(1)$ yields
$$\int_{-\infty}^\infty\frac{\sin(k)}{k}e^{ikx}\,dk=\begin{cases}\pi &,|x|<1\\\\0&,|x|>1\\\\\pi/2&,|x|=1\end{cases} $$
Due to parity it is enough to compute
$$\int_{-\infty}^{+\infty}\frac{\sin^2(x)\cos(xs)}{x^2}\,dx \stackrel{\text{def}}{=}\lim_{M\to +\infty}\int_{-M}^{M}\frac{\sin^2(x)\cos(xs)}{x^2}\,dx $$
and by integration by parts the RHS equals
$$ \lim_{M\to +\infty}\int_{-M}^{M}\frac{\frac{1}{2}\cos(xs)-\frac{1}{4}\cos(x(s+2))-\frac{1}{4}\cos(x(s-2))}{x^2}\,dx $$
or
$$ \lim_{M\to +\infty}\int_{-M}^{M}\frac{\frac{s+2}{4}\sin(x(s+2))+\frac{s-2}{4}\sin(x(s-2))+\frac{s}{2}\sin(xs)}{x}\,dx $$
where we may exploit the standard result
$$ \forall \alpha\in\mathbb{R},\qquad \int_{-\infty}^{+\infty}\frac{\sin(\alpha x)}{x}\,dx = \pi\,\text{Sign}(\alpha) $$
to get:
$$\int_{-\infty}^{+\infty}\frac{\sin^2(x)\cos(xs)}{x^2}\,dx =\frac{\pi}{4}\left[|s-2|+|s+2|-2|s|\right]. $$
The RHS is a piecewise-linear function, supported on $[-2,2]$, going from $0$ to $\pi$ on $[-2,0]$ and from $\pi$ to $0$ on $[0,2]$. Not by chance, it is a multiple of the convolution between $\mathbb{1}_{(-1,1)}$ and itself.
Best Answer
About $(1)$
We make use of Plancherel-Parseval's Theorem: $$\int_{-\infty}^{+\infty} f(x)\overline{g(x)}\text{d}x = \int_{-\infty}^{+\infty} \hat{f(s)} \overline{\hat{g(s)}}\text{d}x$$ Here we can choose for simplicity $$f(x) = \dfrac{\sin(x)}{x}$$ $$g(x) = \dfrac{1}{x^2+1}$$ By the very standard calculus of Fourier Transform you should know that $$\hat{f(s)} = \frac{1}{2} \sqrt{\frac{\pi }{2}} (\text{sgn}(1-s)+\text{sgn}(s+1))$$ and $$\hat{g(s)} = \sqrt{\frac{\pi }{2}} e^{-| s| }$$ whence we get \begin{equation*} \begin{split} \int_{-\infty}^{+\infty} & \left(\frac{1}{2} \sqrt{\frac{\pi }{2}} (\text{sgn}(1-s)+\text{sgn}(s+1))\right)\cdot\left(\sqrt{\frac{\pi }{2}} e^{-| s| }\right)\text{d} s \\\\ & = \dfrac{\pi}{4}\int_{-\infty}^{+\infty}(\text{sgn}(1-s)+\text{sgn}(s+1))e^{-| s| }\text{d} s \\\\ & = \frac{(e-1) \pi }{e} \end{split} \end{equation*}
Remark 1: constants may vary, according to the definition of Fourier Transform used.
If you have doubts about how to compute the Fourier Transforms above, write a comment!
Remark 2: Notice that you could have chosen a different $f$ and $g$, like $f = \sin(x)$ and consequently $g$, and the result would have been the same. Perhaps easy, since F.T. of the sine function gives you two Dirac Delta. You can try it for fun.
About $(2)$
We use the convolution property of the Fourier Transform, that is: $$\mathcal{F}^{-1}(f\cdot g) = \mathcal{F}^{-1}(f) * \mathcal{F}^{-1}(g)$$ with $$f = \dfrac{1 + i\omega}{1 + i\omega}$$ $$g = \dfrac{\sin(\omega)}{\omega}$$ We have respectively: $$\mathcal{F}^{-1}(f) = 2 \sqrt{2 \pi } e^s \theta (-s)-\sqrt{2 \pi } \delta (s)$$ and the other one comes from above: $$\mathcal{F}^{-1}\left(\dfrac{\sin(\omega)}{\omega}\right) = \frac{1}{2} \sqrt{\frac{\pi }{2}} (\text{sgn}(1-s)+\text{sgn}(s+1))$$ Whence the inverse Fourier Transform of their product is give by the convolution of the two transforms: $$\left(2 \sqrt{2 \pi } e^s \theta (-s)-\sqrt{2 \pi } \delta (s)\right) * \left(\frac{1}{2} \sqrt{\frac{\pi }{2}} (\text{sgn}(1-s)+\text{sgn}(s+1))\right)$$ that is $$\dfrac{\pi}{2}\left(2e^s \theta (-s)-\sqrt{2 \pi } \delta (s)\right) * \left(\text{sgn}(1-s)+\text{sgn}(s+1))\right)$$ Assuming you know how to manage convolutions, the final result will be $$4 \left(e^{x+1}-1\right) \theta (-x-1)- 4 \left(e^{x-1}-1\right) \theta (1-x) + (\text{sgn}(1-x)+\text{sgn}(x+1))$$