p-adic congruences motivated by series

Let $p>5$ be a prime. Motivated by the known formulae ${\sum }_{k=1}^{\mathrm{\infty }}(-1{)}^k\mathrm{/}(k^3\left(\genfrac{}{}{0px}{}{2k}{k}\right))=-2\zeta (3)\mathrm{/}5$ and ${\sum }_{k=0}^{\mathrm{\infty }}{\left(\genfrac{}{}{0px}{}{2k}{k}\right)}^2\mathrm{/}((2k+1){16}^k)=4G\mathrm{/}\pi$$(where$G=\sum_{k=0}^\infty(-1)^k/(2k+1)^2$istheCatalanconstant),weshowthat$${\sum }_{k=1}^{(p-1)\mathrm{/}2}\frac{(-1{)}^k}{k^3\left(\genfrac{}{}{0px}{}{2k}{k}\right)}\equiv -2B_{p-3}(\mathrm{m}\mathrm{o}\mathrm{d}p),$$ $$\sum _{k=(p+1)\mathrm{/}2}^{p-1}\frac{{\left(\genfrac{}{}{0px}{}{2k}{k}\right)}^2}{(2k+1){16}^k}\equiv -\frac{7}{4}{p}^2{B}_{p-3}(\mathrm{m}\mathrm{o}\mathrm{d}{p}^3)\text{,}$$ and $$\sum _{k=0}^{(p-3)\mathrm{/}2}\frac{{\left(\genfrac{}{}{0px}{}{2k}{k}\right)}^2}{(2k+1){16}^k}\equiv -2q_p(2)-p{q}_p(2{)}^2+\frac{5}{12}{p}^2{B}_{p-3}(\mathrm{m}\mathrm{o}\mathrm{d}{p}^3),$$ where $B_0,B_1,\dots$ are Bernoulli numbers and $q_p(2)$ is the Fermat quotient $(2^{p-1}-1)\mathrm{/}p$.