The first series has the value
$$\mathcal{S}:=\sum_{n=1}^{\infty}\frac{\left(\psi{\left(\frac{n}{2}\right)}+\frac{1}{n}\right)^2-\left(\psi{\left(\frac{n+1}{2}\right)}\right)^2}{n}=\frac83\ln^3{(2)}-\frac34\,\zeta{(3)}+2\gamma\,\ln^2{(2)}.$$
A useful formula here is the following addition identity for the digamma function:
$$\psi{\left(z+\frac12\right)}+\psi{\left(z\right)}=2\psi{\left(2z\right)}-2\ln{(2)}.$$
For convenience, I define the auxiliary function $\beta{\left(z\right)}$ in terms of the digamma functions as
$$\beta{\left(z\right)}:=\frac12\left[\psi{\left(\frac{z+1}{2}\right)}-
\psi{\left(\frac{z}{2}\right)}\right].$$
(Note: my choice of using $\beta$ to name the function above is simply to be in accordance with the notation adopted in Gradshteyn since it is my primary reference, and it should not be confused with the Dirichlet beta function.)
The beta function can be represented as an integral via
$$\beta{\left(z\right)}=\int_{0}^{1}\frac{t^{z-1}}{1+t}\,\mathrm{d}t;~~~\small{\Re{(z)}>0}.$$
Evaluation of first sum:
$$\begin{align}
\mathcal{S}
&=\sum_{n=1}^{\infty}\frac{\left(\psi{\left(\frac{n}{2}\right)}+\frac{1}{n}\right)^2-\left(\psi{\left(\frac{n+1}{2}\right)}\right)^2}{n}\\
&=\sum_{n=1}^{\infty}\frac{\left[\frac{1}{n}+\psi{\left(\frac{n}{2}\right)}+\psi{\left(\frac{n+1}{2}\right)}\right]\left[\frac{1}{n}-\left(\psi{\left(\frac{n+1}{2}\right)}-\psi{\left(\frac{n}{2}\right)}\right)\right]}{n}\\
&=\sum_{n=1}^{\infty}\frac{\left[\frac{1}{n}+2\psi{\left(n\right)}-2\ln{(2)}\right]\left[\frac{1}{n}-\left(\psi{\left(\frac{n+1}{2}\right)}-\psi{\left(\frac{n}{2}\right)}\right)\right]}{n}\\
&=\sum_{n=1}^{\infty}\frac{\left[\frac{1}{n}+2\psi{\left(n\right)}\right]\left[\frac{1}{n}-\left(\psi{\left(\frac{n+1}{2}\right)}-\psi{\left(\frac{n}{2}\right)}\right)\right]}{n}\\
&~~~~~+2\ln{(2)}\sum_{n=1}^{\infty}\frac{\left[\left(\psi{\left(\frac{n+1}{2}\right)}-\psi{\left(\frac{n}{2}\right)}\right)-\frac{1}{n}\right]}{n}\\
&=\sum_{n=1}^{\infty}\frac{\left[\frac{1}{n}+2\psi{\left(n\right)}\right]\left[\frac{1}{n}-2\beta{\left(n\right)}\right]}{n}+2\ln{(2)}\sum_{n=1}^{\infty}\frac{\left[2\beta{\left(n\right)}-\frac{1}{n}\right]}{n}\\
&=\sum_{n=1}^{\infty}\left[\frac{1}{n^3}+\frac{2(\psi{\left(n\right)}-\beta{\left(n\right)})}{n^2}-\frac{4\psi{\left(n\right)}\beta{\left(n\right)}}{n}\right]+2\ln{(2)}\sum_{n=1}^{\infty}\left[\frac{2\beta{\left(n\right)}}{n}-\frac{1}{n^2}\right]\\
&=\sum_{n=1}^{\infty}\left[\frac{2(\psi{\left(n\right)}-\beta{\left(n\right)})}{n^2}-\frac{4\psi{\left(n\right)}\beta{\left(n\right)}}{n}\right]+4\ln{(2)}\sum_{n=1}^{\infty}\frac{\beta{\left(n\right)}}{n}+\zeta{(3)}-2\ln{(2)}\,\zeta{(2)}\\
&=2\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}-\beta{\left(n\right)}}{n^2}-4\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}\beta{\left(n\right)}}{n}+4\ln{(2)}\sum_{n=1}^{\infty}\frac{\beta{\left(n\right)}}{n}+\zeta{(3)}-2\ln{(2)}\,\zeta{(2)}\\
&=\zeta{(3)}-2\ln{(2)}\,\zeta{(2)}+4\ln{(2)}B_{1}+2D_{2}-2B_{2}-4\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}\beta{\left(n\right)}}{n}\\
&=\zeta{(3)}-2\ln{(2)}\,\zeta{(2)}+4\ln{(2)}B_{1}+2D_{2}-2B_{2}-4\sum_{n=1}^{\infty}\frac{\left[H_{n}-\frac{1}{n}-\gamma\right]\beta{\left(n\right)}}{n}\\
&=\zeta{(3)}-2\ln{(2)}\,\zeta{(2)}+4\ln{(2)}B_{1}+2D_{2}-2B_{2}\\
&~~~~~+4\gamma\sum_{n=1}^{\infty}\frac{\beta{\left(n\right)}}{n}+4\sum_{n=1}^{\infty}\frac{\beta{\left(n\right)}}{n^2}-4\sum_{n=1}^{\infty}\frac{H_{n}\,\beta{\left(n\right)}}{n}\\
&=\zeta{(3)}-2\ln{(2)}\,\zeta{(2)}+4\ln{(2)}B_{1}+2D_{2}-2B_{2}+4\gamma\,B_{1}+4B_{2}-4\sum_{n=1}^{\infty}\frac{H_{n}\,\beta{\left(n\right)}}{n}\\
&=\zeta{(3)}-2\ln{(2)}\,\zeta{(2)}+4\left(\ln{(2)}+\gamma\right)B_{1}+2D_{2}+2B_{2}-4\sum_{n=1}^{\infty}\frac{H_{n}\,\beta{\left(n\right)}}{n},\\
\end{align}$$
where $B_{1},B_{2},D_{2}$ have been introduced for the sake of compactness to stand for the simpler series,
$$\begin{cases}
&B_{1}:=\sum_{n=1}^{\infty}\frac{\beta{\left(n\right)}}{n},\\
&B_{2}:=\sum_{n=1}^{\infty}\frac{\beta{\left(n\right)}}{n^2},\\
&D_{2}:=\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}}{n^2}.\\
\end{cases}$$
First we evaluate the series $\sum_{n=1}^{\infty}\frac{H_{n}\,\beta{\left(n\right)}}{n}$.
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_{n}\,\beta{\left(n\right)}}{n}
&=\sum_{n=1}^{\infty}\frac{H_{n}}{n}\int_{0}^{1}\frac{t^{n-1}}{1+t}\,\mathrm{d}t\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{1+t}\sum_{n=1}^{\infty}\frac{H_{n}\,t^{n-1}}{n}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{1+t}\sum_{n=1}^{\infty}\frac{t^{n-1}}{n}\int_{0}^{1}\mathrm{d}u\,\frac{1-u^{n}}{1-u}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{1+t}\int_{0}^{1}\mathrm{d}u\,\sum_{n=1}^{\infty}\frac{t^{n-1}}{n}\cdot\frac{1-u^{n}}{1-u}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{1+t}\int_{0}^{1}\mathrm{d}u\,\frac{1}{1-u}\sum_{n=1}^{\infty}\frac{t^{n-1}\left(1-u^{n}\right)}{n}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{1+t}\int_{0}^{1}\mathrm{d}u\,\frac{1}{1-u}\left[\frac{\ln{\left(1-tu\right)}-\ln{\left(1-t\right)}}{t}\right]\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{t\left(1+t\right)}\int_{0}^{1}\mathrm{d}u\,\frac{\ln{\left(\frac{1-tu}{1-t}\right)}}{1-u}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{Li}_{2}{\left(t\right)}+\frac12\ln^2{\left(1-t\right)}}{t\left(1+t\right)}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{Li}_{2}{\left(t\right)}+\frac12\ln^2{\left(1-t\right)}}{t}-\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{Li}_{2}{\left(t\right)}+\frac12\ln^2{\left(1-t\right)}}{1+t}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{Li}_{2}{\left(t\right)}}{t}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln^2{\left(1-t\right)}}{2t}\\
&~~~~~-\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{Li}_{2}{\left(t\right)}}{1+t}-\int_{0}^{1}\mathrm{d}t\,\frac{\ln^2{\left(1-t\right)}}{2\left(1+t\right)}\\
&=2\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{Li}_{2}{\left(t\right)}}{t}-\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{Li}_{2}{\left(t\right)}}{1+t}-\int_{0}^{1}\mathrm{d}t\,\frac{\ln^2{\left(1-t\right)}}{2\left(1+t\right)},\\
\end{align}$$
where in the last line above we've made use of the following equality of integrals:
$$\begin{align}
\int_{0}^{1}\mathrm{d}t\,\frac{\ln^2{\left(1-t\right)}}{2t}
&=\left[\frac{\ln{\left(t\right)}\,\ln^2{\left(1-t\right)}}{2}\right]_{0}^{1}-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}\,\ln{\left(1-t\right)}}{t-1}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}\,\ln{\left(1-t\right)}}{1-t}\\
&=\int_{0}^{1}\mathrm{d}u\,\frac{\ln{\left(1-u\right)}\,\ln{\left(u\right)}}{u};~~~\small{1-t=u}\\
&=\left[-\ln{\left(u\right)}\,\operatorname{Li}_{2}{\left(u\right)}\right]_{0}^{1}+\int_{0}^{1}\mathrm{d}u\,\frac{\operatorname{Li}_{2}{\left(u\right)}}{u}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{Li}_{2}{\left(t\right)}}{t}.\\
\end{align}$$
The first of the three integrals to be evaluated is simply $\operatorname{Li}_{3}{\left(1\right)}=\zeta{(3)}$, and the third integral turns out to be equal to $\operatorname{Li}_{3}{\left(\frac12\right)}$:
$$\begin{align}
\int_{0}^{1}\mathrm{d}t\,\frac{\ln^2{\left(1-t\right)}}{2\left(1+t\right)}
&=\int_{\frac12}^{0}\mathrm{d}u\,(-2)\,\frac{\ln^2{\left(2u\right)}}{2\left(2-2u\right)};~~~\small{\left[\frac{1-t}{2}=u\right]}\\
&=\int_{0}^{\frac12}\mathrm{d}u\,\frac{\ln^2{\left(2u\right)}}{2\left(1-u\right)}\\
&=\frac14\int_{0}^{1}\mathrm{d}x\,\frac{\ln^2{\left(x\right)}}{1-\left(\frac12\right)x};~~~\small{\left[2u=x\right]}\\
&=\frac14\cdot\frac{2\operatorname{Li}_{3}{\left(z\right)}}{z}\bigg{|}_{z=\frac12}\\
&=\operatorname{Li}_{3}{\left(\frac12\right)}.\\
\end{align}$$
The remaining integral may be evaluated as follows:
$$\begin{align}
\int_{0}^{1}\frac{\operatorname{Li}_{2}{\left(t\right)}}{1+t}\,\mathrm{d}t
&=\left[\ln{\left(1+t\right)}\,\operatorname{Li}_{2}{\left(t\right)}\right]_{0}^{1}+\int_{0}^{1}\frac{\ln{\left(1-t\right)}\,\ln{\left(t+1\right)}}{t}\,\mathrm{d}t\\
&=\ln{(2)}\,\zeta{(2)}+\int_{0}^{1}\frac{\ln{\left(1-t\right)}\,\ln{\left(t+1\right)}}{t}\,\mathrm{d}t\\
&=\ln{(2)}\,\zeta{(2)}+\int_{0}^{1}\frac{\ln^2{\left(1-t\right)}+\ln^2{\left(1+t\right)}-\ln^2{\left(\frac{1+t}{1-t}\right)}}{2t}\,\mathrm{d}t\\
&=\ln{(2)}\,\zeta{(2)}+\int_{0}^{1}\frac{\ln^2{\left(1-t\right)}}{2t}\,\mathrm{d}t+\int_{0}^{1}\frac{\ln^2{\left(1+t\right)}}{2t}\,\mathrm{d}t-\int_{0}^{1}\frac{\ln^2{\left(\frac{1+t}{1-t}\right)}}{2t}\,\mathrm{d}t\\
&=\ln{(2)}\,\zeta{(2)}+\zeta{(3)}+\frac18\zeta{(3)}-\int_{0}^{1}\frac{\ln^2{\left(\frac{1+t}{1-t}\right)}}{2t}\,\mathrm{d}t\\
&=\ln{(2)}\,\zeta{(2)}+\frac98\zeta{(3)}-\int_{0}^{1}\frac{\ln^2{\left(u\right)}}{1-u^2}\,\mathrm{d}u;~~~\small{\frac{1-t}{1+t}=u}\\
&=\ln{(2)}\,\zeta{(2)}+\frac98\zeta{(3)}-\frac74\zeta{(3)}\\
&=\ln{(2)}\,\zeta{(2)}-\frac58\zeta{(3)}.\\
\end{align}$$
$$\begin{align}
B_{1}
&=\sum_{n=1}^{\infty}\frac{\beta{\left(n\right)}}{n}\\
&=\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1}\mathrm{d}t\,\frac{t^{n-1}}{t+1}\\
&=\int_{0}^{1}\mathrm{d}t\,\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{t^{n-1}}{t+1}\right)\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{t+1}\sum_{n=1}^{\infty}\frac{t^{n-1}}{n}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{t\left(t+1\right)}\sum_{n=1}^{\infty}\frac{t^{n}}{n}\\
&=\int_{0}^{1}\frac{(-1)\ln{\left(1-t\right)}}{t\left(t+1\right)}\,\mathrm{d}t\\
&=-\int_{0}^{1}\frac{\ln{\left(1-t\right)}}{t}\,\mathrm{d}t+\int_{0}^{1}\frac{\ln{\left(1-t\right)}}{t+1}\,\mathrm{d}t\\
&=\zeta{(2)}+\int_{0}^{1}\frac{\ln{\left(1-t\right)}}{t+1}\,\mathrm{d}t\\
&=\zeta{(2)}+\int_{0}^{1}\frac{\ln{\left(u\right)}}{2-u}\,\mathrm{d}u;~~~\small{\left[1-t=u\right]}\\
&=\zeta{(2)}+\frac12\ln^2{(2)}-\frac12\,\zeta{(2)}\\
&=\frac12\,\zeta{(2)}+\frac12\ln^2{(2)}.\\
\end{align}$$
$$\begin{align}
B_{2}
&=\sum_{n=1}^{\infty}\frac{\beta{\left(n\right)}}{n^2}\\
&=\sum_{n=1}^{\infty}\frac{1}{n^2}\int_{0}^{1}\mathrm{d}t\,\frac{t^{n-1}}{t+1}\\
&=\int_{0}^{1}\mathrm{d}t\,\sum_{n=1}^{\infty}\frac{1}{n^2}\left(\frac{t^{n-1}}{t+1}\right)\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{t+1}\sum_{n=1}^{\infty}\frac{t^{n-1}}{n^2}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{t\left(t+1\right)}\sum_{n=1}^{\infty}\frac{t^{n}}{n^2}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{1}{t\left(t+1\right)}\,\operatorname{Li}_{2}{\left(t\right)}\\
&=\int_{0}^{1}\frac{\operatorname{Li}_{2}{\left(t\right)}}{t\left(t+1\right)}\,\mathrm{d}t\\
&=\int_{0}^{1}\frac{\operatorname{Li}_{2}{\left(t\right)}}{t}\,\mathrm{d}t-\int_{0}^{1}\frac{\operatorname{Li}_{2}{\left(t\right)}}{t+1}\,\mathrm{d}t\\
&=\zeta{(3)}-I\\
&=\frac{13}{8}\,\zeta{(3)}-\ln{(2)}\,\zeta{(2)}.\\
\end{align}$$
$D_{2}\approx0.25245=\zeta{(3)}-\gamma\,\zeta{(2)}$
$$\begin{align}
D_{2}
&=\sum_{n=1}^{\infty}\frac{\psi{\left(n\right)}}{n^2}\\
&=\sum_{n=1}^{\infty}\frac{1}{n^2}\left[-\gamma+\int_{0}^{1}\frac{t-t^{n}}{t-t^2}\,\mathrm{d}t\right]\\
&=-\gamma\sum_{n=1}^{\infty}\frac{1}{n^2}+\sum_{n=1}^{\infty}\frac{1}{n^2}\int_{0}^{1}\mathrm{d}t\,\frac{t-t^{n}}{t-t^2}\\
&=-\gamma\,\zeta{(2)}+\int_{0}^{1}\mathrm{d}t\,\sum_{n=1}^{\infty}\frac{1}{n^2}\left(\frac{t-t^{n}}{t-t^2}\right)\\
&=-\gamma\,\zeta{(2)}+\int_{0}^{1}\mathrm{d}t\,\frac{1}{t-t^2}\sum_{n=1}^{\infty}\frac{t-t^{n}}{n^2}\\
&=-\gamma\,\zeta{(2)}+\int_{0}^{1}\mathrm{d}t\,\frac{1}{t-t^2}\left[\zeta{(2)}\,t-\operatorname{Li}_{2}{\left(t\right)}\right]\\
&=-\gamma\,\zeta{(2)}+\int_{0}^{1}\mathrm{d}t\,\frac{\zeta{(2)}-\frac{\operatorname{Li}_{2}{\left(t\right)}}{t}}{1-t}\\
&=\zeta{(3)}-\gamma\,\zeta{(2)}.\\
\end{align}$$
Now it's just a matter of adding up all the specified terms, upon which one finds the final value stated at the top of this response.
