Finally, following the excellent suggestion of an anonymous referee,
recalling that
$$h_{k} = H_{2k} - \frac{1}{2}\,H_{k},\eqno(29)$$
we find from (19)
$$\sum_{k=1}^{\infty}\,\frac{1}{k^{2}}\,H_{2k} = \sum_{k=1}^{\infty}\,\frac{1}{k^{2}}\,h_{k}
+ \frac{1}{2}\,\sum_{k=1}^{\infty}\,\frac{1}{k^{2}}\,H_{k} =
\frac{11}{4}\,\zeta(3).\eqno(30)$$ Furthermore, in terms of the
multiple series [7]
$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\,\frac{1}{ij(i + j)} =
2\zeta(3),\,\,\,\,\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\,\frac{(-1)^{i
+ j}}{ij(i + j)} = \frac{1}{4}\,\zeta(3),$$ the difference gives
$$\sum_{i, j> 1, i + j = \mbox{odd}}\,\frac{1}{ij(i + j)} =
\frac{7}{8}\,\zeta(3).$$ Setting $i + j = 2k + 1$ and using partail
fractions, we have
\begin{eqnarray*}
\sum_{i, j> 1, i + j = \mbox{odd}}\,\frac{1}{ij(i + j)} & = &
\sum_{k=1}^{\infty}\,\sum_{j=1}^{2k}\,\frac{1}{j(2k+1-j)(2k + 1)}\\
& = & \sum_{k=1}^{\infty}\,\frac{1}{(2k+ 1)^{2}}\sum_{j=1}^{2k}\,\left(\frac{1}{j} + \frac{1}{2k+1-j}\right)\\
& = & \sum_{k=1}^{\infty}\,\frac{1}{(2k+ 1)^{2}}\,2H_{2k}.
\end{eqnarray*}
Thus,
$$\sum_{k=1}^{\infty}\,\frac{1}{(2k+ 1)^{2}}\,H_{2k} =
\frac{7}{16}\,\zeta(3).\eqno(31)$$  Subsequently, we have
$$\sum_{k=1}^{\infty}\,\frac{1}{(2k-1)^{2}}\,H_{2k}
  =  \sum_{k=1}^{\infty}\,\frac{1}{(2k+ 1)^{2}}\,H_{2k} + $$
$$\sum_{k=1}^{\infty}\,\frac{1}{(2k- 1)^{3}}
 + \sum_{k=1}^{\infty}\,\frac{1}{2k(2k- 1)^{2}}
 = \frac{21}{16}\,\zeta(3) +
 \frac{1}{8}\,(\pi^{2} - 8\ln 2).\eqno(32)$$
From this and the known result [1]
$$\sum_{k=1}^{\infty}\,\frac{1}{(2k- 1)^{2}}\,H_{k} =
\frac{1}{4}\,(\pi^{2} - \pi^{2}\ln 2 - 8\ln 2 + 7\zeta(3)),$$ we
finally get
$$\sum_{k=1}^{\infty}\,\frac{1}{(2k- 1)^{2}}\,h_{k} =
\frac{7}{16}\,\zeta(3) + \frac{3}{4}\,\zeta(2)\ln 2. \eqno(33)$$