2) $$(\frac{a+4}{a-4} - \frac{a-4}{a+4}):\frac{48a}{16-a^2}.$$

2) $$(\frac{a+4}{a-4} - \frac{a-4}{a+4}):\frac{48a}{16-a^2} = (\frac{(a+4)^2 - (a-4)^2}{(a-4)(a+4)}) : \frac{48a}{-(a^2-16)} = (\frac{a^2+8a+16 - (a^2-8a+16)}{(a-4)(a+4)}) \cdot \frac{-(a^2-16)}{48a} = \frac{16a}{(a-4)(a+4)} \cdot \frac{-(a-4)(a+4)}{48a} = \frac{16a \cdot (-1)}{48a} = -\frac{1}{3}$$

Ответ: -$$\frac{1}{3}$$