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
$$\int_{-\infty}^\infty\frac{\sin(k)}{k}e^{ikx}\,dk=\begin{cases}\pi &,|x|<1\\\\0&,|x|>1\\\\\pi/2&,|x|=1\end{cases} $$
There is a much simpler way by using the Fourier inversion theorem!
Then the result just follows from the fact that
$$
{\cal F}^{-1}(\mathbf 1_{[-1,1]})(x) = \frac{1}{2\pi}\int_{-1}^1 e^{i\,\omega\,x}\,\mathrm d \omega = \frac{e^{ix}-e^{-ix}}{2i\pi\,x} = \frac{\sin x}{\pi\,x}.
$$
and so
$$
{\cal F}\left(\frac{\sin x}{x}\right) = \pi\,{\cal F}\left(\frac{\sin x}{\pi\,x}\right) = \pi\,\mathbf 1_{[-1,1]}(\omega)
$$
Let me add details if you are not familiar with the Fourier transform:
the Fourier inversion theorem tells that you can find the function from its Fourier transform by writing
$$
f(x) = \frac{1}{2\pi} \int_{\Bbb R} \widehat f(\omega) \, e^{i\omega\,x}\,\mathrm d \omega = \frac{1}{2\pi} \cal F\big(\hat f\big)(-x)
$$
where $\widehat f = {\cal F}(f)$. Equivalently, it means that the inverse operation to the Fourier transform is given by
$$
{\cal F}^{-1}(g)(x) = \frac{1}{2\pi} \int_{\Bbb R} g(\omega) \, e^{i\omega\,x}\,\mathrm d \omega = \frac{1}{2\pi} \cal F(g)(-x)
$$
Best Answer
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.