2) a²-4a a4-64a; 36a²-1 36a²-120+1

2) $$\frac{a^2 - 4a}{36a^2 - 1} : \frac{a^4 - 64a}{36a^2 - 12a + 1} = \frac{a(a-4)}{(6a-1)(6a+1)} : \frac{a(a^3 - 64)}{(6a-1)^2} = \frac{a(a-4)}{(6a-1)(6a+1)} \cdot \frac{(6a-1)^2}{a(a^3 - 64)} = \frac{a(a-4)(6a-1)^2}{a(6a-1)(6a+1)(a^3 - 64)} = \frac{(a-4)(6a-1)}{(6a+1)(a^3 - 64)} = \frac{(a-4)(6a-1)}{(6a+1)(a-4)(a^2 + 4a + 16)} = \frac{6a-1}{(6a+1)(a^2 + 4a + 16)}$$

Ответ: $$\frac{6a-1}{(6a+1)(a^2 + 4a + 16)}$$