In order to clarify the four cases - and possibly correction in the question - we have:
$$ \sum_{n=1}^{\infty}\frac{\eta(2n+1)}{2^{2n+1}} \quad = \frac{1}{2}\int_{0}^{\infty}\frac{\cosh(x/2)-1}{e^{x}+1}\,dx = \frac{1}{2}\left(1-\log2\right) \tag{1} \\[6mm] $$
$$ \sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{2^{2n+1}} \quad = \frac{1}{2}\int_{0}^{\infty}\frac{\cosh(x/2)-1}{e^{x}-1}\,dx = \frac{1}{2}\left(2\log2-1\right) \tag{2} \\[6mm] $$
$$ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{\eta(2n+1)}{2^{2n+1}} = \frac{1}{2}\int_{0}^{\infty}\frac{1-\cos(x/2)}{e^{x}+1}\,dx = \frac{1}{2}\log2\,+ \qquad\quad\qquad \frac{1}{8}\left(H_{-1/2+i/4}+H_{-1/2-i/4}\right)-\frac{1}{8}\left(H_{+i/4}+H_{-i/4}\right) \tag{3} \\[6mm] $$
$$ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{\zeta(2n+1)}{2^{2n+1}} = \frac{1}{2}\int_{0}^{\infty}\frac{1-\cos(x/2)}{e^{x}-1}\,dx = \frac{1}{4}\left(H_{+i/2}+H_{-i/2}\right) \tag{4} \\[8mm] $$
$$ \small \color{blue}{H_s=\gamma+\psi(s+1)=\int_{0}^{1}\frac{1-x^s}{1-x}\,dx}\quad\text{"Generalized Harmonic Number"} $$
$$
\#\space\sum_{n=1}^{\infty}\frac{\eta(2n+1)}{2^{2n+1}} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{\Gamma(2n+1)\,\eta(2n+1)}{\Gamma(2n+1)\,2^{2n}} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{2^{2n}\,(2n)!}\int_{0}^{\infty}\frac{x^{2n}}{e^x+1}\,dx = \\
\frac{1}{2}\int_{0}^{\infty}\frac{1}{e^x+1}\left(\sum_{n=1}^{\infty}\frac{(x/2)^{2n}}{(2n)!}\right)\,dx = \frac{1}{2}\int_{0}^{\infty}\frac{\cosh(x/2)-1}{e^{x}+1}\,dx = \color{red}{\frac{1}{2}\left(1-\log2\right)} \\[8mm]
$$
$$
\#\space\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\zeta(2n+1)}{2^{2n+1}} = \frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\Gamma(2n+1)\,\zeta(2n+1)}{\Gamma(2n+1)\,2^{2n}} = \\
\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{2^{2n}\,(2n)!}\int_{0}^{\infty}\frac{x^{2n}}{e^x-1}\,dx = \frac{1}{2}\int_{0}^{\infty}\frac{1}{e^x-1}\left(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(x/2)^{2n}}{(2n)!}\right)\,dx = \\
\frac{1}{2}\int_{0}^{\infty}\frac{1-\cos(x/2)}{e^{x}-1}\,dx = \color{red}{\frac{1}{4}\left(H_{+i/2}+H_{-i/2}\right)} = \frac{1}{2}\,\Re(H_{i/2}) = \frac{1}{2}\left[\gamma+\Re\left(\psi(1+i/2)\right)\right] \\[8mm]
$$
And, unfortunately, neither $Re\left\{H_{i/2}\right\}$ nor $Re\left\{\psi(1+i/2)\right\}$ have known closed form in term of log, exp or similar functions.