New series for $\zeta(5)$ involving second-order harmonic numbers

Question

In 1997 T. Amdeberhan and D. Zeilberger proved that $$\sum_{k=1}^\infty\frac{(-1)^k(205k^2-160k+32)}{k^5\binom{2k}k^5}=-2\zeta(3).\tag{1}$$ In 2008 J. Guillera obtained that $$\sum_{k=1}^\infty\frac{(10k^2-6k+1)(-256)^k}{k^5\binom{2k}k^5}=-28\zeta(3).\tag{2}$$ Recently K. C. Au [arXiv:2212.02986] confirmed the identity $$\sum_{k=1}^\infty\frac{(28k^2-18k+3)(-64)^k}{k^5\binom{2k}k^4\binom{3k}k}=-14\zeta(3)\tag{3}$$ conjectured by me in 2010.

Motivated by $(1)-(3)$, my recent preprint and Question 436205, I made the following conjecture on new series for $\zeta(5)$ involving second-order harmonic numbers $$H_n^{(2)}:=\sum_{0<k\le n}\frac1{k^2}\ \ \ \ (n=0,1,2,3,\ldots).$$

Conjecture 1. We have \begin{align}\sum_{k=1}^\infty\frac{(-1)^k((205k^2-160k+32)(4H_{2k-1}^{(2)}-12H_{k-1}^{(2)})-43)}{k^5\binom{2k}k^5}&=-8\zeta(5),\tag4 \\\sum_{k=1}^\infty\frac{(-256)^k((10k^2-6k+1)(4H_{2k-1}^{(2)}-3H_{k-1}^{(2)})-2)}{k^5\binom{2k}k^5}&=-124\zeta(5),\tag5 \\\sum_{k=1}^\infty\frac{(-64)^k((28k^2-18k+3)(2H_{2k-1}^{(2)}-3H_{k-1}^{(2)})-2)}{k^5\binom{2k}k^4\binom{3k}k}&=-31\zeta(5).\tag6 \end{align}

Remark. In the same spirit, motivated by the known identities \begin{align}\sum_{k=0}^\infty(20k^2+8k+1)\frac{\binom{2k}k^5}{(-4096)^k}&=\frac 8{\pi^2},\tag7 \\\sum_{k=0}^\infty(74k^2+27k+3)\frac{\binom{2k}k^4\binom{3k}k}{4096^k}&=\frac{48}{\pi^2},\tag8 \\\sum_{k=0}^\infty(120k^2+34k+3)\frac{\binom{2k}k^4\binom{4k}{2k}}{2^{16k}}&=\frac{32}{\pi^2}\tag{9} \\\sum_{k=0}^\infty(820k^2+180k+13)\frac{\binom{2k}k^5}{(-2^{20})^k}&=\frac{128}{\pi^2}\tag{10} \end{align} (cf. arXiv:2101.12592), we conjecture the identities \begin{align}\sum_{k=0}^\infty\frac{\binom{2k}k^5}{(-4096)^k}((20k^2+8k+1)(8H_{2k}^{(2)}-3H_k^{(2)})+4)&=\frac 83,\tag{11} \\\sum_{k=0}^\infty\frac{\binom{2k}k^4\binom{3k}k}{4096^k}((74k^2+27k+3)(92H_{2k}^{(2)}-33H_k^{(2)})+112)&=272,\tag{12} \\\sum_{k=0}^\infty\frac{\binom{2k}k^4\binom{4k}{2k}}{2^{16k}}((120k^2+34k+3)(23H_{2k}^{(2)}-7H_k^{(2)})+24)&=\frac{88}3.\tag{13} \\\sum_{k=0}^\infty\frac{\binom{2k}k^5}{(-2^{20})^k}((820k^2+180k+13)(11H_{2k}^{(2)}-3H_k^{(2)})+43)&=\frac{128}3.\tag{14} \end{align}

Motivated by $(7)-(9)$ and Question 436205, we also formulate the following conjecture on series for $(\log 2)/\pi^2$ involving the harmonic numbers $H_n=\sum_{0<k\le n}\frac 1k\ (n=0,1,2,\ldots)$.

Conjecture 2. We have \begin{align}\sum_{k=0}^\infty\frac{\binom{2k}k^5}{(-4096)^k}((20k^2+8k+1)(15H_{2k}-5H_k)+2)&=\frac{16\log2}{\pi^2},\tag{15} \\\sum_{k=0}^\infty\frac{\binom{2k}k^4\binom{3k}k}{4096^k}((74k^2+27k+3)(51H_{3k}+250H_{2k}-153H_k)+15)&=\frac{9792\log2}{\pi^2},\tag{16} \\\sum_{k=0}^\infty\frac{\binom{2k}k^4\binom{4k}{2k}}{2^{16k}}((120k^2+34k+3)(34H_{4k}+4H_{2k}-8H_k)+19)&=\frac{512\log2}{\pi^2}.\tag{17} \end{align}

Actually we have many other similar conjectural series.

QUESTION. Are the identities $(4)-(6)$ and $(11)-(17)$ true? Can one find a way to prove some of them?

Your comments are welcome!