Let $\phi$ be the famous golden ratio $\frac{\sqrt5+1}2$, and let $\zeta(3)=\sum_{n=1}^\infty\frac1{n^3}.$

In 2014, I formulated the following conjecture involving both $\phi$ and $\zeta(3)$.

Conjecture. We have the identities $$\sum_{k=1}^\infty\frac{L_{2k}}{k^2\binom{2k}k}\left(\frac1k+\frac1{k+1}+\cdots+\frac1{2k}\right)=\frac{41\zeta(3)+4\pi^2\log\phi}{25}\tag{1}$$ and $$\sum_{k=1}^\infty\frac{v_k}{k^2\binom{2k}k}\left(\frac1k+\frac1{k+1}+\cdots+\frac1{2k}\right)=\frac{124\zeta(3)+\pi^2\log\left(5^5\phi^6\right)}{50}, \tag{2}$$ where the Lucas numbers $L_0,L_1,L_2,\ldots$ are given by $$L_0=2,\ L_1=1,\ \text{and}\ L_{n+1}=L_n+L_{n-1}\ \ \text{for all}\ n=1,2,3,\ldots,$$ and $v_0,v_1,v_2,\ldots,$ are defined by $$v_0=2,\ v_1=5,\ \text{and}\ v_{n+1}=5(v_n-v_{n-1})\ \ \text{for all}\ n=1,2,3,\ldots.$$

Remark. I found $(1)$ and $(2)$ on Nov. 29, 2014 and Dec. 7, 2014 respectively. As the two series converge rapidly, we can easily check $(1)$ and $(2)$ numerically. It is easy to show that $$\sum_{k=1}^\infty\frac{L_{2k}}{k^2\binom{2k}k}=\frac{\pi^2}5\ \ \text{and}\ \ \sum_{k=1}^\infty\frac{v_k}{k^2\binom{2k}k}=\frac25\pi^2.$$

QUESTION. How to prove the identities $(1)$ and $(2)$?