Is there a closed-form of $\frac{\zeta (2)}{\pi ^2}+\frac{\zeta (4)}{\pi ^4}+\frac{\zeta (6)}{\pi ^6}+.....$

$$\begin{align} \sum_{n=1}^{\infty}\frac{\zeta(2n)}{\pi^{2n}} &=\sum_{n=1}^{\infty}\frac{1}{\pi^{2n}}\sum_{k=1}^{\infty}\frac{1}{k^{2n}}\tag{1}\\ &=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{(\pi^2 k^2)^n}\tag{2}\\ &=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{(\pi^2 k^2)^n}\tag{3}\\ &=\sum_{k=1}^{\infty}\frac{1}{(\pi^2 k^2)-1}\tag{4}\\ &=\frac{1}{\pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2-\frac1{\pi^2}}\tag{5}\\ &=\frac{1}{2\pi}\left[\pi-\pi\cot\left(\frac\pi\pi\right)\right]\tag{6}\\ &=\frac{1-\cot1}{2}\tag{7} \end{align}$$

$$\sum_{n=1}^{\infty}\frac{\zeta(2n)}{\pi^{2n}}=\frac{1-\cot1}{2}$$

$\text{Explanation :}$ $\text{6}$ using

$$\sum_{k=1}^\infty\frac{1}{k^2-z^2} =\frac{1}{2z^2}-\frac{\pi\cot(\pi z)}{2z} =\frac{1}{2z}\left[\frac1z-\pi\cot(\pi z)\right]$$

Recalling that $$\zeta\left(2n\right)=\left(-1\right)^{n+1}\frac{B_{2n}2^{2n-1}\pi^{2n}}{\left(2n\right)!},\ n\geq1$$the Laurent expansion of $\cot$ function at the origin in terms of the Riemann zeta function is $$\cot\left(z\right)=\frac{1}{z}-2\sum_{k\geq1}\zeta\left(2k\right)\frac{z^{2k-1}}{\pi^{2k}},\ 0<\left|z\right|<\pi$$ so for your series take $z=1$.