\(a=\sqrt{5}\left(\sqrt{6}+1\right):\sqrt{\frac{\left(2\sqrt{3}+\sqrt{2}\right)^2}{\left(2\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+\sqrt{2}\right)}}=\sqrt{5}\left(\sqrt{6}+1\right):\frac{\left(2\sqrt{3}+\sqrt{2}\right)}{\sqrt{10}}\)
\(=\sqrt{5}\left(\sqrt{6}+1\right).\frac{\sqrt{5}.\sqrt{2}}{\left(\sqrt{6}+1\right).\sqrt{2}}=5\)
\(b=\frac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{\left(\sqrt{2}-\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}+\frac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2}+\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{-\sqrt{3}}+\frac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{\sqrt{3}}=\frac{1}{\sqrt{3}}\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)\)
\(=\frac{1}{\sqrt{6}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)=\frac{1}{\sqrt{6}}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\right)\)
\(=\frac{1}{\sqrt{6}}\left(\sqrt{3}+1+\sqrt{3}-1\right)=\frac{2\sqrt{3}}{\sqrt{6}}=\sqrt{2}\)