The proof of (1), sketched in http://www.maths.lancs.ac.uk/~jameson/catalan.pdf, goes as follows.

With the substitution $x=x^\prime/2,\,y=y^\prime/2$, the integral (1) takes the form (we have omitted the primes)
$$I_2=\int_0^1\int_0^1\frac{dx\,dy}{4-x^2-y^2}.$$
Due to the $x\leftrightarrow y$ exchange symmetry of the integrand,
$$I_2=2\iint\limits_\Delta \frac{dx\,dy}{4-x^2-y^2},$$
where the triangular integration domain $\Delta$ is the lower half of the unit square $0\le x\le 1,\,0\le y\le 1$. Let us introduce the polar coordinates, as suggested by Aeryk, $x=r\cos{\theta},\,y=r\sin{\theta}$. Then
$$I_2=\int\limits_0^{\pi/4}d\theta\int\limits_0^{1/cos{\theta}}\frac{2r}{4-r^2}dr=\int\limits_0^{\pi/4}\ln{\frac{4\cos^2{\theta}}{4\cos^2{\theta}-1}}\,d\theta.$$
Now, by using $\sin{3\theta}=\sin{\theta}(3\cos^2{\theta}-\sin^2{\theta})=\sin{\theta}(4\cos^2{\theta}-1)$, we cen rewrite the above integral in the form $$I_2=\int\limits_0^{\pi/4}\ln{\frac{4\cos^2{\theta}\sin{\theta}}{\sin{3\theta}}}\,d\theta=\frac{\pi}{2}\ln{2}+2I_C+I_S-I_{3S},$$
where the integrals $$I_S=\int\limits_0^{\pi/4}\ln{\sin{\theta}}\,d\theta=-\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2},\;I_C=\int\limits_0^{\pi/4}\ln{\cos{\theta}}\,d\theta=\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2}$$
were calculated in http://www.maths.lancs.ac.uk/~jameson/catalan.pdf, while for $I_{3S}$ we have
$$I_{3S}=\int\limits_0^{\pi/4}\ln{\sin{3\theta}}\,d\theta=\frac{1}{3}
\int\limits_0^{3\pi/4}\ln{\sin{\theta}}\,d\theta=\frac{1}{3}\left[
\int\limits_0^{\pi/2}\ln{\sin{\theta}}\,d\theta+
\int\limits_{\pi/2}^{3\pi/4}\ln{\sin{\theta}}\,d\theta\right].$$
The substitution $\alpha=-(\pi/2-\theta)$ shows that the second integral equals to $I_C$, while for the first integral the following result can be found in http://www.maths.lancs.ac.uk/~jameson/catalan.pdf
$$\int\limits_0^{\pi/2}\ln{\sin{\theta}}\,d\theta=
\int\limits_0^{\pi/2}\ln{\cos{\theta}}\,d\theta=-\frac{\pi}{2}\,\ln{2}.$$
Therefore
$$I_{3S}=\frac{1}{3}\left[-\frac{\pi}{2}\,\ln{2}+\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2}\right]=\frac{1}{6}\,G-\frac{\pi}{4}\,\ln{2},$$ and
$$I_2=\frac{\pi}{2}\ln{2}+2\left(\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2}\right)+\left(-\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2}\right)-\left(\frac{1}{6}\,G-\frac{\pi}{4}\,\ln{2}\right)=\frac{G}{3}.$$
By the similar method, the following result was proved in http://www.maths.lancs.ac.uk/~jameson/catalan.pdf
$$I_1=\int_0^1\int_0^1\frac{dx\,dy}{2-x^2-y^2}=G.$$