I am looking for a short proof that ∫0∞(sin⁡xx)2dx=π2 What do you think?

Well, it's not hard to reduce this integral to ${\displaystyle f\left(x\right)=max\{0,1-\left|x\right|\}}$. It is easy to calculate the Fourier transform
${\displaystyle \hat{f}\left(\xi \right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(x\right)e^{-ix\xi }={\left(\frac{\sin \left(\frac{\xi }{2}\right)}{\frac{\xi }{2}}\right)}^2}$
Taking the inverse Fourier transform, we get
${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left(\frac{\sin \left(\frac{\xi }{2}\right)}{\frac{\xi }{2}}\right)}^2{e}^{ix\xi }d\xi =2\pi f\left(x\right)}$,
and the result follows.
The second integral can be computed in a similar way. Just take ${\displaystyle f\left(x\right)={\chi }_{\begin{array}{cc}-1& 1\end{array}}\left(x\right)}$ (the indicator function of the interval ${\displaystyle [-1,1])}$

${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2}x}{x^2}dx={\int }_0^{\mathrm{\infty }}\frac{\frac{12}{1-\cos 2x}}{x^2}dx}$
${\displaystyle ={\int }_0^{\mathrm{\infty }}\frac{1-\cos x}{x^2}dx}$
${\displaystyle =\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1-\cos x}{x^2}dx}$
${\displaystyle ={\frac{12}{\int }}_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1-\cos x}{x^2}dx}$
${\displaystyle =\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathfrak{R}\frac{1-e\left\{ix\right\}}{x^2}dx}$
${\displaystyle =\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathfrak{R}\frac{1-e^{ix}+i\frac{x}{1+x^2}}{x^2}dx}$
${\displaystyle =\frac{1}{2}\mathfrak{R}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1-e^{ix}+i\frac{x}{1+x^2}}{x^2}dx}$
Now close the contour in the upper half plane, enclosing the pole at ${\displaystyle x=i}$ with residue ${\displaystyle \frac{1}{2i}}$, yielding
${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2}x}{x^2}dx=\frac{1}{2}\cdot 2\pi \cdot \frac{1}{2i}=\frac{\pi }{2}}$

From squaring the identity
$\frac{\sin nx}{\sin x}=\frac{e^{inx}-e^{-inx}}{e^{ix}-e-ix}=\sum _{k=0}^{n-1}{e}^{(2k-n+1)ix}$
and integrating we get
$n\pi ={\int }_{-\pi /2}^{\pi /2}\frac{{\sin }^{2}nx}{{\sin }^{2}x}dx$
Let
$I_n={\int }_{-\pi /2}^{\pi /2}\frac{{\sin }^{2}nx}{n{x}^2}dx={\int }_{-n\pi /2}^{n\pi /2}\frac{{\sin }^{2}y}{y^2}dy$
Then
$\pi -I_n=\frac{1}{n}{\int }_{-\pi /2}^{\pi /2}{\sin }^{2}nx({\csc }^{2}x-x^{-2x})dx$
and so
$|\pi -I_n|\le \frac{1}{n}{\int }_{-\pi /2}^{\pi /2}|{\csc }^{2}x-x^{-2}|dx=O(1/n)$
as $x\mapsto {\csc }^{2}x-x^{-2}$ extends to a continuous function on $[-\pi /2,\pi /2]$. Hence $I_n\to \pi$ as $n\to \mathrm{\infty }$ and
$\pi ={\int }_{\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\sin }^{2}y}{y^2}dy$