A hypergeometric series for $\sqrt3\pi$ with converging rate $1/9$

Question

Recently, I found a (conjectural) new series for $\sqrt3\pi$: $$\sum_{k=1}^\infty\frac{(8k-3)\binom{4k}{2k}}{k(4k-1)9^k\binom{2k}k^2}=\frac{\sqrt3\pi}{18}.\label{1}\tag{1}$$ The series converges fast with converging rate $1/9$, so one can easily check the identity \eqref{1} numerically.

Motivated by \eqref{1}, I also conjecture that $$\sum_{k=1}^\infty\frac{\binom{4k}{2k}\left((8k-3)(5H_{2k-1}-4H_{k-1})-6\right)}{k(4k-1)9^k\binom{2k}k^2}=\frac32\sum_{n=0}^\infty\left(\frac1{(3n+1)^2}-\frac1{(3n+2)^2}\right) \label{2}\tag{2}$$ and $$\sum_{k=1}^\infty\frac{\binom{4k}{2k}(8k-3)\left(2H_{2k-1}^{(2)}-5H_{k-1}^{(2)}\right)}{k(4k-1)9^k\binom{2k}k^2}=\frac{\pi^3}{36\sqrt3}, \label{3}\tag{3}$$ where $$H_n:=\sum_{0<k\le n}\frac1k\ \ \ \text{and}\ \ \ \ H_n^{(2)}:=\sum_{0<k\le n}\frac1{k^2}$$ for each $n=0,1,2,\ldots$.

QUESTION. Can one prove the new identities \eqref{1},\eqref{2} and \eqref{3} by using known tools (such as the WZ method and various hypergeometric series identities) ?

For the identities \eqref{1},\eqref{2} and \eqref{3}, we also have corresponding conjectural $p$-adic congruences. For example, I conjecture that for any prime $p>3$ we have $$\sum_{k=1}^{(p-1)/2}\frac{(8k-3)\binom{4k}{2k}}{k(4k-1)9^k\binom{2k}k^2}\equiv-\frac5{36}pB_{p-2}\left(\frac13\right)\pmod{p^2}\tag{4}$$ and $$\sum_{k=1}^{(p-1)/2}\frac{\binom{4k}{2k}\left((8k-3)(5H_{2k-1}-4H_{k-1})-6\right)}{k(4k-1)9^k\binom{2k}k^2}\equiv\frac16B_{p-2}\left(\frac13\right)\pmod p,\tag{5}$$ where $B_{p-2}(x)$ denotes the Bernoulli polynomial of degree $p-2$.

Your comments are welcome!