(1)

\begin{align*} H_{\frac{1}{2},z} & =\zeta\left(z\right)-\sum_{k=1}^{\infty}\frac{1}{\left(k+\frac{1}{2}\right)^{z}}\\ & =\zeta\left(z\right)-2^{z}\sum_{k=1}^{\infty}\frac{1}{\left(2k+1\right)^{z}}\\ & =\zeta\left(z\right)-2^{z}\sum_{k=0}^{\infty}\frac{1}{\left(2k+1\right)^{z}}+2^{z}\\ & =\zeta\left(z\right)-2^{z}\frac{2^{z}-1}{2^{z}}\zeta\left(z\right)+2^{z}\\ & =2^{z}-\left(2^{z}-2\right)\zeta\left(z\right) \end{align*}

(2)

\begin{align*} H_{\frac{1}{4},2} & =\zeta\left(2\right)-\sum_{k=1}^{\infty}\frac{1}{\left(k+\frac{1}{4}\right)^{2}}\\ & =\zeta\left(2\right)-4^{2}\sum_{k=1}^{\infty}\frac{1}{\left(4k+1\right)^{2}}\\ & =\zeta\left(2\right)-4^{2}\sum_{k=0}^{\infty}\frac{1}{\left(4k+1\right)^{2}}+4^{2}\\ & =\zeta\left(2\right)-4^{2}\frac{1}{2}\left(\sum_{k=0}^{\infty}\frac{1}{\left(2k+1\right)^{2}}+\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{2}}\right)+4^{2}\\ & =\zeta\left(2\right)-4^{2}\frac{1}{2}\left(\frac{2^{2}-1}{2^{2}}\zeta\left(2\right)+G\right)+4^{2}\\ & =16-5\zeta\left(2\right)-8G\\ & =16-\frac{5}{6}\pi^{2}-8G \end{align*}

(3)

\begin{align*} H_{\frac{3}{4},2} & =\zeta\left(2\right)-\sum_{k=1}^{\infty}\frac{1}{\left(k+\frac{3}{4}\right)^{2}}\\ & =\zeta\left(2\right)-4^{2}\sum_{k=1}^{\infty}\frac{1}{\left(4k+3\right)^{2}}\\ & =\zeta\left(2\right)-4^{2}\sum_{k=0}^{\infty}\frac{1}{\left(4k+3\right)^{2}}+\left(\frac{4}{3}\right)^{2}\\ & =\zeta\left(2\right)-4^{2}\frac{1}{2}\left(\sum_{k=0}^{\infty}\frac{1}{\left(2k+3\right)^{2}}+\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+3\right)^{2}}\right)+\left(\frac{4}{3}\right)^{2}\\ & =\zeta\left(2\right)-4^{2}\frac{1}{2}\left(\sum_{k=0}^{\infty}\frac{1}{\left(2k+1\right)^{2}}-1-\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{2}}+1\right)+\left(\frac{4}{3}\right)^{2}\\ & =\zeta\left(2\right)-4^{2}\frac{1}{2}\left(\frac{2^{2}-1}{2^{2}}\zeta\left(2\right)-G\right)+\left(\frac{4}{3}\right)^{2}\\ & =\frac{16}{9}-5\zeta\left(2\right)+8G\\ & =\frac{16}{9}-\frac{5}{6}\pi^{2}+8G \end{align*}

(4)

\begin{align*} H_{\frac{1}{4},3} & =\zeta\left(3\right)-\sum_{k=1}^{\infty}\frac{1}{\left(k+\frac{1}{4}\right)^{3}}\\ & =\zeta\left(3\right)-4^{3}\sum_{k=1}^{\infty}\frac{1}{\left(4k+1\right)^{3}}\\ & =\zeta\left(3\right)-4^{3}\sum_{k=0}^{\infty}\frac{1}{\left(4k+1\right)^{3}}+4^{3}\\ & =\zeta\left(3\right)-4^{3}\frac{1}{2}\left(\sum_{k=0}^{\infty}\frac{1}{\left(2k+1\right)^{3}}+\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{3}}\right)+4^{3}\\ & =\zeta\left(3\right)-4^{3}\frac{1}{2}\left(\frac{2^{3}-1}{2^{3}}\zeta\left(3\right)+\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{3}}\right)+4^{3}\\ & =4^{3}-\left(4^{3}\frac{2^{3}-1}{2^{4}}-1\right)\zeta\left(3\right)-4^{3}\frac{1}{2}\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{3}}\\ & =64-27\zeta\left(3\right)-\pi^{3}\cmt{\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{3}}=\frac{\pi^{3}}{32}} \end{align*}

(5)

\begin{align*} H_{\frac{3}{4},3} & =\zeta\left(3\right)-\sum_{k=1}^{\infty}\frac{1}{\left(k+\frac{3}{4}\right)^{3}}\\ & =\zeta\left(3\right)-4^{3}\sum_{k=1}^{\infty}\frac{1}{\left(4k+3\right)^{3}}\\ & =\zeta\left(3\right)-4^{3}\sum_{k=0}^{\infty}\frac{1}{\left(4k+3\right)^{3}}+\left(\frac{4}{3}\right)^{3}\\ & =\zeta\left(3\right)-4^{3}\frac{1}{2}\left(\sum_{k=0}^{\infty}\frac{1}{\left(2k+3\right)^{3}}+\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+3\right)^{3}}\right)+\left(\frac{4}{3}\right)^{3}\\ & =\zeta\left(3\right)-4^{3}\frac{1}{2}\left(\sum_{k=0}^{\infty}\frac{1}{\left(2k+1\right)^{3}}-1-\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{3}}+1\right)+\left(\frac{4}{3}\right)^{3}\\ & =\zeta\left(3\right)-4^{3}\frac{1}{2}\left(\frac{2^{3}-1}{2^{3}}\zeta\left(3\right)-\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{3}}\right)+\left(\frac{4}{3}\right)^{3}\\ & =\left(\frac{4}{3}\right)^{3}-\left(4^{3}\frac{2^{3}-1}{2^{4}}-1\right)\zeta\left(3\right)+4^{3}\frac{1}{2}\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{3}}\\ & =\left(\frac{4}{3}\right)^{3}-27\zeta\left(3\right)+\pi^{3}\cmt{\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{3}}=\frac{\pi^{3}}{32}} \end{align*}