a)
{\frac{9+6\sqrt5+5-[2^2-(\sqrt5)^2]}{(\sqrt7)^2-2^2}=\frac{14+6\sqrt5-4+5}{7-4}=\frac{15+6\sqrt5}{3}=\frac{3(5+2\sqrt5)}{3}=5+2\sqrt5}
b)
{\frac{(1+\sqrt{12})^2-(2-\sqrt3)(\sqrt3-2)}{(\sqrt8-2)(2+2\sqrt2)}=\frac{1+2\sqrt{4*3}+(\sqrt{12})^2+(\sqrt3-2)^2}{(\sqrt{4*2}-2)(2+2\sqrt2)}=}
=\frac{1+2*2\sqrt3+12+3-4\sqrt3+4}{(2\sqrt2-2)(2\sqrt2+2)}=\frac{20}{4*2-4}=\frac{20}{4}=5
c)
{(\sqrt{3-\sqrt5}+\sqrt{\sqrt5+3})^2=3-\sqrt5+2*\sqrt{3^2-(\sqrt5)^2}+\sqrt5+3=6+2\sqrt{9-5}=6+2\sqrt4=6+2*2=10}
d)
{(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3})^2=2+\sqrt3-2\sqrt{(2+\sqrt3)(2-\sqrt3)}+2-\sqrt3=2-2\sqrt{4-3}+2=2-2*1+2=2}