Proving that $\left(\frac{\pi}{2}\right)^{2}=1+\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k-1}}$. Wolfram$\alpha$ says that we have the following identity
$$
\left(\frac{\pi}{2}\right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k}}
$$

but, how does one prove such identity?
It's well known that
$$
\pi\cot(\pi x)=\frac 1x-2\sum_{n=1}^\infty\zeta(2n)x^{2n-1};
$$
taking derivatives we get
$$
\frac{\pi^2}{\sin^2\pi x}=x^{-2}+2\sum_{n=1}^\infty(2n-1)\zeta(2n)x^{2n-2};
$$
in particular for $x=\frac12$
$$
\pi^2=4+2\sum_{n=1}^\infty(2n-1)\zeta(2n)\frac1{2^{2n-2}};
$$
this is your formula (multiplied by 4).
Do not worry about the downvote. It is just an attack on my answers.

The series can have the following integral representation


  $$ \sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k-1}} = \int_{0}^{1}\frac{(t^2+1)\ln t}{t^2-1}  dt = \frac{\pi^2}{4}-1.$$