How can we prove the following identities $$\sum_{n=1}^\infty n^{-3}{\binom{2n}n}^{-1}=\pi\operatorname{Cl}_2\left(\frac{2\pi}{3}\right)-\frac{4}{3}\zeta(3)\tag{1}$$ $$\sum_{n=1}^\infty (n+1)^{-2}{\binom{2n}n}^{-1}=\frac{2\pi^2}{9}-2\pi\operatorname{Cl}_2\left(\frac{2\pi}{3}\right)+\frac{8}{3}\zeta(3)-1\tag{2}$$ where $\operatorname{Cl}_2(x)$ is the Clausen integral?
Some identities of this sort are proved in this paper.
$(1)$: We want to evaluate $\quad\displaystyle S:=\sum_{m=1}^{\infty} \frac{1}{m^3\binom{2m}{m}}$
From this answer we obtained : $$\tag{1}2(\arcsin(x))^2=\sum_{m=1}^{\infty} \frac{(2x)^{2m}}{m^2\binom{2m}{m}}$$ Integrating this multiplied by $\,\dfrac 2x\,$ from $\,0$ to $\dfrac 12$ will thus give : \begin{align} S&=2\sum_{m=1}^{\infty} \frac{1}{m^2\binom{2m}{m}}\int_0^{\frac12}\dfrac{(2x)^{2m}}{x}\,dx\\ \tag{2}S&=4\int_0^{\frac 12} \frac{(\arcsin(x))^2}x\,dx\\ \tag{3}S&=4\int_0^{\pi/6} \frac{t^2}{\tan(t)}\,dt\\ S&=4\left[\left.t^2\log(\sin(t))\right|_{\,0}^{\,\pi/6}-2\int_0^{\pi/6} t\,\log(\sin(t))\,dt\right]\\ \tag{4}S&=-\frac{\log(2)\pi^2}9-8\int_0^{\pi/6} t\,\log(\sin(t))\,dt\\ \end{align} The Clausen integral verifies : $\;\displaystyle\operatorname{Cl}_2(x)'=-\log(2\sin(t/2))\;$ so let's rewrite $(4)$ and use integration by parts of $\operatorname{Cl}_2(x)$ : \begin{align} S&=-\frac{\log(2)\pi^2}9-\frac 84\int_0^{\pi/3} t\;\log(2\sin(t/2))-t\log(2)\,dt\\ \tag{5}S&=-2\int_0^{\pi/3} t\;\log(2\sin(t/2))\,dt\\ &=2\left[t\;\operatorname{Cl}_2(x)\left.\right|_0^{\pi/3}-\int_0^{\pi/3} \operatorname{Cl}_2(t)\,dt\right]\\ &=\frac{2\pi}3\operatorname{Cl}_2\left(\frac{\pi}3\right)+2\,\left(\operatorname{Cl}_3\left(\frac{\pi}3\right)-\operatorname{Cl}_3\left(0\right)\right)\\ \end{align} Since $\;\displaystyle\operatorname{Cl}_{2n}(x):=\sum_{k=1}^\infty \frac{\sin(k\,x)}{k^{\,2n}},\ \operatorname{Cl}_{2n+1}(x):=\sum_{k=1}^\infty \frac{\cos(k\,x)}{k^{\,2n+1}}\;$ we have indeed $\;\operatorname{Cl}_3(x)'=-\operatorname{Cl}_2(x)$.
Now $\,\operatorname{Cl}_3(0)=\zeta(3)\,$ and $\,\operatorname{Cl}_3\left(\dfrac{\pi}3\right)=\dfrac{\zeta(3)}3\,$ (prove this using the series for $\operatorname{Cl}_3$) while $\,\operatorname{Cl}_2\left(\dfrac{2\pi}3\right)=\dfrac 23\operatorname{Cl}_2\left(\dfrac{\pi}3\right)\,$ can't be written in simpler form (without using polylogarithms) so that your $(1)$ is indeed right : $$\boxed{\displaystyle\sum_{n=1}^\infty \frac 1{n^{3}{\binom{2n}n}}=\pi\operatorname{Cl}_2\left(\frac{\pi}{3}\right)-\frac{4}{3}\zeta(3)}\tag{6}$$
$(2)$: Concerning $\;\displaystyle \sum_{m=1}^{\infty} \frac{1}{(m+1)^2\binom{2m}{m}}\ $ the link (i.e. the derivative of $(1)$ multiplied by $\dfrac x2$) gives us : $$\tag{7}\frac{2x \arcsin\ x}{\sqrt{1-x^2}}=\sum_{m=1}^{\infty} \frac{(2x)^{2m}}{m\binom{2m}{m}}$$ The derivative of this (multiplied by $x/2$) will give us the general :
$$\tag{8}\boxed{\frac {x^2}{1-x^2}+x\frac {\arcsin(x)}{\sqrt{1-x^2}^3}=\sum_{m=1}^{\infty} \frac{(2x)^{2m}}{\binom{2m}{m}}}$$
Multiplying by $\,x$, integrating and dividing by $x^2/2$ we get :
$$\tag{9}2\frac {\arcsin(x)}{x\,\sqrt{1-x^2}}-\frac{\arcsin(x)^2}{x^2}-1=\sum_{m=1}^{\infty} \frac{(2x)^{2m}}{(m+1)\binom{2m}{m}}$$
The indefinite integral of $\;\displaystyle 2\;x\frac {\arcsin(x)}{x\,\sqrt{1-x^2}}\;$ is simply $\,\arcsin(x)^2\,$ while the integral of $\;\displaystyle x\frac{\arcsin(x)^2}{x^2}$ is more complicated but we found $\;\displaystyle\int_0^{1/2}\frac {\arcsin(x)^2}x\,dx=\frac S4$ earlier and can therefore conclude that : $$\int_0^{1/2} \sum_{m=1}^{\infty} \frac{x(2x)^{2m}}{(m+1)\binom{2m}{m}}\,dx=\sum_{m=1}^{\infty} \frac{2^{-3}(1)^{2m+2}}{(m+1)^2\binom{2m}{m}}=\left.\arcsin(x)^2-\frac{x^2}2\right|_{\,0}^{\,1/2}-\frac S4$$ or : $$\sum_{m=1}^{\infty} \frac 1{(m+1)^2\binom{2m}{m}}=8\left(\frac{\pi}6\right)^2-\frac 88-2\left(\pi\operatorname{Cl}_2\left(\frac{\pi}{3}\right)-\frac{4}{3}\zeta(3)\right)$$ which is indeed your equality $(2)$ : $$\tag{10}\boxed{\displaystyle\sum_{m=1}^{\infty} \frac 1{(m+1)^2\binom{2m}{m}}=\frac{2\pi^2}{9}-2\pi\operatorname{Cl}_2\left(\frac{2\pi}{3}\right)+\frac{8}{3}\zeta(3)-1}$$
To add to the links provided by Vladimir Reshetnikov :