(1)

\begin{align*} \sum_{k=1}^{\infty}\left(\zeta(4k)-1\right) & =\frac{1}{2}\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)\left(1^{k}+\left(-1\right)^{k}\right)\\ & =\frac{1}{2}\left(\left[\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)x^{2k}\right]_{x=1}+\left[\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)x^{2k}\right]_{x=i}\right)\\ & =\frac{1}{2}\left(\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)+\left[\frac{1}{2}\left(1-\pi x\tan^{-1}\left(\pi x\right)\right)-\frac{x^{2}}{1-x^{2}}\right]_{x=i}\right)\cmt{\sum_{k=1}^{\infty}\zeta(2k)x^{2k}=\frac{1}{2}\left(1-\pi x\tan^{-1}\left(\pi x\right)\right)}\\ & =\frac{1}{2}\left(\frac{3}{4}+\frac{1}{2}\left(1-\pi\tanh^{-1}\pi\right)+\frac{1}{2}\right)\cmt{\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)=\frac{3}{4}}\\ & =\frac{7}{8}-\frac{\pi}{4}\tanh^{-1}\pi \end{align*}

(2)

\begin{align*} \sum_{k=1}^{\infty}\left(\zeta(4k-2)-1\right) & =\frac{1}{2}\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)\left(1^{k}-\left(-1\right)^{k}\right)\\ & =\frac{1}{2}\left(\left[\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)x^{2k}\right]_{x=1}-\left[\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)x^{2k}\right]_{x=i}\right)\\ & =\frac{1}{2}\left(\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)-\left[\frac{1}{2}\left(1-\pi x\tan^{-1}\left(\pi x\right)\right)-\frac{x^{2}}{1-x^{2}}\right]_{x=i}\right)\cmt{\sum_{k=1}^{\infty}\zeta(2k)x^{2k}=\frac{1}{2}\left(1-\pi x\tan^{-1}\left(\pi x\right)\right)}\\ & =\frac{1}{2}\left(\frac{3}{4}-\frac{1}{2}\left(1-\pi\tanh^{-1}\pi\right)-\frac{1}{2}\right)\cmt{\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)=\frac{3}{4}}\\ & =-\frac{1}{8}+\frac{\pi}{4}\tanh^{-1}\pi \end{align*}