It seems to have escaped attention that these sums may be evaluated
using harmonic summation techniques.
Introduce the sum
$$S(x; \alpha, p) = \sum_{n\ge 1} \frac{1}{n^p(e^{\alpha n x}-1)}$$
with $p$ a positive odd integer and $\alpha>1$, so that we seek e.g.
$2 S(1; \pi\sqrt{2}, 3)+S(1; 2\pi\sqrt{2}, 3).$
The sum term is harmonic and may be evaluated by inverting its Mellin
transform.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = \frac{1}{k^p}, \quad \mu_k = k
\quad \text{and} \quad
g(x) = \frac{1}{e^{\alpha x}-1}.$$
We need the Mellin transform $g^*(s)$ of $g(x)$ which is
$$\int_0^\infty \frac{1}{e^{\alpha x}-1} x^{s-1} dx
= \int_0^\infty \frac{e^{-\alpha x}}{1-e^{-\alpha x}} x^{s-1} dx
\\ = \int_0^\infty \sum_{q\ge 1} e^{-\alpha q x} x^{s-1} dx
= \sum_{q\ge 1} \int_0^\infty e^{-\alpha q x} x^{s-1} dx
\\= \Gamma(s) \sum_{q\ge 1} \frac{1}{(\alpha q)^s}
= \frac{1}{\alpha^s} \Gamma(s) \zeta(s).$$
It follows that the Mellin transform $Q(s)$ of the harmonic sum
$S(x;\alpha,p)$ is given by
$$Q(s) = \frac{1}{\alpha^s} \Gamma(s) \zeta(s) \zeta(s+p)
\quad\text{because}\quad
\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} =
\sum_{k\ge 1} \frac{1}{k^p} \frac{1}{k^s}
= \zeta(s+p)$$
for $\Re(s) > 1-p.$
The Mellin inversion integral here is
$$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$
which we evaluate by shifting it to the left for an expansion about
zero.
First formula.
We take
$$Q(s) = \frac{1}{\pi^s\sqrt{2}^s}
\left(2 + \frac{1}{2^s}\right)
\Gamma(s) \zeta(s) \zeta(s+3).$$
We shift the Mellin inversion integral to the line $s=-1$, integrating
right through the pole at $s=-1$ picking up the following residues:
$$\mathrm{Res}(Q(s)/x^s; s=1) =
\frac{\pi^3\sqrt{2}}{72 x}
\quad\text{and}\quad
\mathrm{Res}(Q(s)/x^s; s=0) =
-\frac{3}{2}\zeta(3)$$
and
$$\frac{1}{2}\mathrm{Res}(Q(s)/x^s; s=-1) =
\frac{\pi^3\sqrt{2}x}{36}.$$
This almost concludes the proof of the first formula if we can show
that the integral on the line $\Re(s) = -1$ vanishes when $x=1.$ To
accomplish this we must show that the integrand is odd on this line.
Put $s = -1 - it$ in the integrand to get
$$\pi^{1+it} \sqrt{2}^{1+it}
\left(2 + 2^{1+it}\right)
\Gamma(-1-it) \zeta(-1-it) \zeta(2-it).$$
Now use the functional equation of the Riemann Zeta function in the
following form:
$$\zeta(1-s) = \frac{2}{2^s\pi^s}
\cos\left(\frac{\pi s}{2}\right) \Gamma(s) \zeta(s)$$
to obtain (with $s=-1-it$)
$$\pi^{1+it} \sqrt{2}^{1+it}
\left(2 + 2^{1+it}\right)
\zeta(2+it) 2^{-1-it} \pi^{-1-it}
\frac{1}{2\cos\left(\frac{\pi (-1-it)}{2}\right)}
\zeta(2-it)$$
which is
$$ \sqrt{2}^{1+it}
\left(2^{-it} + 1\right)
\zeta(2+it)
\frac{1}{2\cos\left(\frac{\pi (1+it)}{2}\right)}
\zeta(2-it)$$
and finally yields
$$-\frac{1}{\sin(\pi i t/2)}
\left(\sqrt{2}^{-1-it}+\sqrt{2}^{-1+it}\right)
\zeta(2+it)\zeta(2-it).$$
It is now possible to conclude by inspection: the zeta function terms
and the powers of the square root are even in $t$ and the sine term is
odd, so the whole term is odd and the integral vanishes. (We get exponential decay from the sine term.)
Second formula.
We take
$$Q(s) = \frac{1}{\pi^s\sqrt{2}^s}
\left(4 - \frac{1}{2^s}\right)
\Gamma(s) \zeta(s) \zeta(s+5).$$
We shift the Mellin inversion integral to the line $s=-2$ (no pole on
the line this time) picking up the following residues:
$$\mathrm{Res}(Q(s)/x^s; s=1) =
\frac{\pi^5\sqrt{2}}{540 x}
\quad\text{and}\quad
\mathrm{Res}(Q(s)/x^s; s=0) =
-\frac{3}{2}\zeta(5)$$
and
$$\mathrm{Res}(Q(s)/x^s; s=-1) =
\frac{\pi^5\sqrt{2}x}{540}.$$
It remains to verify that the integrand on the line $\Re(s)=-2$ is odd
when $x=1$. Put $s=-2-it$ in the integrand to get
$$\pi^{2+it} \sqrt{2}^{2+it}
\left(4 - 2^{2+it}\right)
\Gamma(-2-it) \zeta(-2-it) \zeta(3-it).$$
Applying the functional equation once again with $s=-2-it$ we obtain
$$\pi^{2+it} \sqrt{2}^{2+it}
\left(4 - 2^{2+it}\right)
\zeta(3+it) 2^{-2-it} \pi^{-2-it}
\frac{1}{2\cos\left(\frac{\pi(-2-it)}{2}\right)}
\zeta(3-it)$$
which is
$$\sqrt{2}^{2+it}
\left(2^{-it} - 1\right)
\zeta(3+it)
\frac{1}{2\cos\left(\frac{\pi(-2-it)}{2}\right)}
\zeta(3-it)$$
which is in turn
$$\left(\sqrt{2}^{2-it} - \sqrt{2}^{2+it} \right)
\frac{1}{2\cos\left(\frac{\pi(2+it)}{2}\right)}
\zeta(3+it)
\zeta(3-it)$$
which finally yields
$$-\left(\sqrt{2}^{2-it} - \sqrt{2}^{2+it} \right)
\frac{1}{2\cos(\pi i t /2)}
\zeta(3+it)
\zeta(3-it)$$
The product of the zeta function terms is even, as is the cosine
term. The term in front is odd, so the integrand is odd as claimed.
(We get exponential decay from the cosine term.)
Third formula.
We take
$$Q(s) = \frac{1}{\pi^s\sqrt{2}^s}
\left(8 + \frac{1}{2^s}\right)
\Gamma(s) \zeta(s) \zeta(s+7).$$
We shift the Mellin inversion integral to the line $s=-3$, integrating
right through the pole at $s=-3$ picking up the following residues:
$$\mathrm{Res}(Q(s)/x^s; s=1) =
\frac{17 \pi^7\sqrt{2}}{37800 x}
\quad\text{and}\quad
\mathrm{Res}(Q(s)/x^s; s=0) =
-\frac{9}{2}\zeta(7)$$
and
$$\mathrm{Res}(Q(s)/x^s; s=-1) =
\frac{\pi^7\sqrt{2}x}{1134}
\quad\text{and}\quad
\frac{1}{2} \mathrm{Res}(Q(s)/x^s; s=-3) =
-\frac{\pi^7\sqrt{2}x^3}{4050}.$$
This almost concludes the proof of this third formula if we can show
that the integral on the line $\Re(s) = -3$ vanishes when $x=1.$ To
accomplish this we must show once more that the integrand is odd on
this line.
Put $s = -3 - it$ in the integrand to get
$$\pi^{3+it} \sqrt{2}^{3+it}
\left(8 + 2^{3+it}\right)
\Gamma(-3-it) \zeta(-3-it) \zeta(4-it).$$
By the functional equation we obtain with $s = -3-it$
$$\pi^{3+it} \sqrt{2}^{3+it}
\left(8 + 2^{3+it}\right)
\zeta(4+it)
2^{-3-it} \pi^{-3-it}
\frac{1}{2\cos\left(\pi(-3-it)/2\right)}
\zeta(4-it)$$
which is
$$\sqrt{2}^{3+it}
\left(2^{-it} + 1\right)
\zeta(4+it)
\frac{1}{2\cos\left(\pi(3+it)/2\right)}
\zeta(4-it)$$
which finally yields
$$\frac{1}{2\sin(\pi it/2)}
\left(\sqrt{2}^{3-it} + \sqrt{2}^{3+it}\right)
\zeta(4+it)
\zeta(4-it).$$
This concludes it since the two zeta function terms together are even
as is the square root term while the sine term is odd, so their
product is odd.
A similar yet not quite the same computation can be found at this MSE link.
Another computation in the same spirit is at this MSE link II.
Here is my solution:
Step 1. Reduction to an integral representation
Let $S$ denote the summation in question. Then by the successive application of integration by parts, we obtain
\begin{align*}
S &= 8 \int_{0}^{1} \frac{1}{x} \int_{0}^{x/2} \frac{\arcsin^{2} t}{t} \, dt dx
= -8\int_{0}^{1} \frac{\arcsin^{2} (x/2)}{x} \log x \, dx \\
&= 8 \int_{0}^{1} \frac{\arcsin (x/2)}{\sqrt{1 - (x/2)^{2}}} \log^{2} x \, \frac{dx}{2}.
\end{align*}
Thus with the substitution $x = 2 \sin\theta$, we have
$$ S = 8 \int_{0}^{\frac{\pi}{6}} \theta \log^{2} (2 \sin\theta) \, d\theta. $$
To evaluate this integral, note that for $0 < \theta < \frac{\pi}{6}$ we have
$$ e^{i\theta} \cdot 2 \sin \theta = i \cdot (1 - e^{2i\theta}). $$
Taking logarithm (with the branch cut $(-\infty, 0]$ as usual) to both sides, it follows that
$$ i\theta + \log (2\sin\theta) = \frac{i\pi}{2} + \log(1 - e^{2i\theta}). $$
Cubing both sides and integrating on $\left( 0, \frac{\pi}{6} \right)$ and taking imaginary parts only,
$$ S = \frac{2}{3} \left( \frac{\pi}{6} \right)^{4} + \frac{8}{3} \Im \int_{0}^{\frac{\pi}{6}} \left( \frac{i\pi}{2} + \log(1 - e^{2i\theta}) \right)^{3} \, d\theta. \tag{1} $$
Step 2. Some complex-analysis techniques
Now we focus on the integral in the imaginary part:
$$ I := \int_{0}^{\frac{\pi}{6}} \left( \frac{i\pi}{2} + \log(1 - e^{2i\theta}) \right)^{3} \, d\theta \tag{2}. $$
Once we evaluate the imaginary part of $I$, the identity $(1)$ immediately gives us the answer. We first make the substitution $z = 1 - e^{2i\theta}$ and $\omega = e^{-i\pi/3}$ to obtain
$$ I = \int_{0}^{\omega} \left( \frac{i\pi}{2} + \log z \right)^{3} \frac{dz}{2i(z-1)}. $$
Here, the path of integration is a circular arc joining from $0$ to $\omega$ centered at $1$ (green-colored path in the figure below).
But since the integrand is analytic for $0 < \Re z < 1$, we may change the path of integration as $z = \omega t$ for $0 \leq t \leq 1$ (blue-colored path in the figure above). This gives
$$ I = \frac{1}{2i} \int_{0}^{1} \left( \frac{i\pi}{6} + \log t \right)^{3} \frac{\omega \, dt}{\omega t - 1}. $$
Plugging $t = e^{-x}$, $I$ reduces to
\begin{align*}
I
&= \frac{1}{2} \int_{0}^{\infty} \left( \frac{\pi}{6} + ix \right)^{3} \frac{\omega e^{-x}}{1 - \omega e^{-x}} \, dx
= \frac{1}{2} \sum_{n=1}^{\infty} \omega^{n} \int_{0}^{\infty} \left( \frac{\pi}{6} + ix \right)^{3} e^{-nx} \, dx \\
&= \frac{1}{2} \sum_{n=1}^{\infty} \omega^{n} \left( -\frac{6 i}{n^4}-\frac{\pi }{n^3}+\frac{i \pi ^2}{12 n^2}+\frac{\pi ^3}{216 n} \right).
\end{align*}
Taking the imaginary part,
\begin{align*}
\Im I
&= -3 \sum_{n=1}^{\infty} \frac{\cos (n\pi/3)}{n^4} + \frac{\pi}{2} \sum_{n=1}^{\infty} \frac{\sin (n\pi/3)}{n^3} + \frac{\pi^2}{24} \sum_{n=1}^{\infty} \frac{\cos (n\pi/3)}{n^2} - \frac{\pi^3}{432} \sum_{n=1}^{\infty} \frac{\sin (n\pi/3)}{n}. \tag{3}
\end{align*}
Step 3. Evaluation of a series
Note that for $0 < \theta < \pi$, we have
$$ \sum_{n=1}^{\infty} \frac{\sin n\theta}{n} = \Im \sum_{n=1}^{\infty} \frac{e^{in\theta}}{n} = - \Im \log(1 - e^{i\theta}) = \frac{\pi-\theta}{2}. $$
Integrating both sides, we obtain
$$ \sum_{n=1}^{\infty} \frac{1-\cos n\theta}{n^{2}} = \frac{\theta (2 \pi -\theta )}{4}
\quad \Longrightarrow \quad
\sum_{n=1}^{\infty} \frac{\cos n\theta}{n^{2}} = \frac{\theta ^2}{4}-\frac{\pi \theta }{2}+\frac{\pi ^2}{6}.$$
Repeating this procedure, we obtain
$$ \sum_{n=1}^{\infty} \frac{\sin n\theta}{n^{3}} = \frac{\theta ^3}{12}-\frac{\pi \theta ^2}{4}+\frac{\pi ^2 \theta }{6}$$
and
$$ \sum_{n=1}^{\infty} \frac{\cos n\theta}{n^{4}} = -\frac{\theta ^4}{48}+\frac{\pi \theta ^3}{12}-\frac{\pi ^2 \theta ^2}{12}+\frac{\pi^4}{90}.$$
Plugging $\theta = \frac{\pi}{3}$, we have
\begin{align*}
\sum_{n=1}^{\infty} \frac{\sin (n \pi / 3)}{n^{2}} &= \frac{\pi}{3} \\
\sum_{n=1}^{\infty} \frac{\cos (n \pi / 3)}{n^{2}} &= \frac{\pi^3}{36} \\
\sum_{n=1}^{\infty} \frac{\sin (n \pi / 3)}{n^{3}} &= \frac{5 \pi^3}{162} \\
\sum_{n=1}^{\infty} \frac{\cos (n \pi / 3)}{n^{4}} &= \frac{91 \pi ^4}{19440}
\end{align*}
Plugging these to $(3)$, we have
$$ \Im I = \frac{23 \pi^4}{12960} \quad \Longrightarrow \quad S = \frac{17 \pi^4}{3240} = \frac{17}{36}\zeta(4) $$
as desired.
Best Answer
The paper gives (second-to-last page, right-hand column)
$$\sum_{n\ge1}\frac{1}{n^5\binom{2n}{n}}=\frac{9\sqrt{3}\pi}{8}\color{Purple}{L\left(4,\left(-3\atop\circ\right)\right)}+\frac{\pi^2\zeta(3)}{9}-\frac{19\zeta(5)}{3}. \tag{1}$$
You seek a way to write the $L$-function in purple as a linear combination of Hurwitz $\zeta$ functions.
More generally, let $\chi$ be a Dirichlet character of modulus (period) $m$, and define a 'Kronecker' delta
$$\delta_m(k)=\begin{cases}1 & k\equiv0\bmod m \\ 0 & k\not\equiv 0\bmod m\end{cases}. \tag{2}$$
Notice then that $\delta_m(a-b)$ is $1$ if and only if $a\equiv b\bmod m$. We can therefore decompose $\chi$ as
$$\chi(n)=\sum_{k=0}^{m-1} \chi(k) \delta_m(n-k). \tag{3}$$
Furthermore, the Hurwitz zeta function at $a/m\in[0,1)$ decomposes as
$$\begin{array}{c l} \zeta\left(s,\frac{a}{m}\right) & =\sum_{n=1}^\infty\frac{1}{(n+a/m)^s} \\ & =m^s\sum_{n=1}^\infty\frac{1}{(mn+a)^s} \\ & =m^s\sum_{n\ge1} \frac{\delta_m(n-a)}{n^s}.\end{array} \tag{4}$$
Therefore, we have
$$\begin{array}{c l} L(s,\chi) & =\sum_{n\ge1}\frac{\chi(n)}{n^s} \\ & =\sum_{n\ge1}\frac{1}{n^s}\sum_{k=0}^{m-1}\chi(k)\delta_m(n-k) \\ & =\sum_{k=0}^{m-1}\chi(k)\sum_{n\ge1}\frac{\delta_m(n-k)}{n^s} \\ & =\frac{1}{m^s}\sum_{k=0}^{m-1}\chi(k)\zeta\left(s,\frac{k}{m}\right). \end{array} \tag{5}$$
This formula is listed on Wikipedia's Hurwitz $\zeta$ and Dirichlet $L$-function articles. In particular,
$$L\left(4,\left(\frac{-3}{\circ}\right)\right)=\frac{\zeta\left(4,\frac{1}{3}\right)-\zeta\left(4,\frac{2}{3}\right)}{81} \tag{6}$$
because $\left(\frac{-3}{1}\right)=1$ and $\left(\frac{-3}{2}\right)=-1$ (and $\chi(0)=0$ for all Dirichlet characters). Also see here.