Даю 50 баллов!!!!!!!!!!!!!!

Решите задачу:

$1.\\a)\;5\sqrt{48}-0,2\sqrt{75}+\frac23\sqrt{27}=5\sqrt{16\cdot3}-0,2\sqrt{25\cdot3}+\frac23\sqrt{9\cdot3}=\\=20\sqrt3-\sqrt3+\sqrt3=20\sqrt3\\b)(\sqrt6-\sqrt2)^2+\sqrt{48}=6-2\sqrt{12}+2-\sqrt{4\cdot12}=8-2\sqrt{12}+2\sqrt{12}=8\\\\2.\\a)\;\frac1{a^2-4}\cdot\frac{(a+2)^2}{a}=\frac1{(a+3)(a+2)}{(a+2)^2}{a}=\frac{a+2}{a(a-1)}\\b)\;\frac{a-3}{a+3}\cdot\frac{a^2+27}{a^2-9}=\frac{a-3}{a+3}\cdot\frac{(a+3)(a^2-3a+9)}{(a-3)(a+3)}=\frac{a^2+27}{(a+3)^2}$