A new series for $\pi^3$ and related congruences

Let $H_n^{(2)}$ denote the second-order harmonic number $\sum_{0<k\le n}1/k^2$ for $n=0,1,2,\ldots$. In this paper we obtain the following identity: $$\sum_{k=1}^\infty\frac{2^kH_{k-1}^{(2)}}{k\binom{2k}k}=\frac{\pi^3}{48}.$$ We explain how we found the series and develop related congruences involving Bernoulli or Euler numbers; for example, it is shown that $$\sum_{k=1}^{p-1}\frac{\binom{2k}k}{2^k}H_k^{(2)}\equiv-E_{p-3}\pmod{p}$$ for any prime $p>3$, where $E_0,E_1,E_2,\ldots$ are Euler numbers. Motivated by the Amdeberhan-Zeilberger identity $\sum_{k=1}^\infty(21k-8)/(k^3\binom{2k}k^3)=\pi^2/6$, we also establish the congruence $$\sum_{k=1}^{(p-1)/2}\frac{21k-8}{k^3\binom{2k}k^3}\equiv(-1)^{(p+1)/2}4E_{p-3}\pmod p$$ for each prime $p>3$.