$$\left(\frac{a+2}{a-2} + \frac{a-2}{a+2}\right) : \frac{16a}{a^2-4} = \left(\frac{(a+2)^2 + (a-2)^2}{(a-2)(a+2)}\right) : \frac{16a}{a^2-4} = \frac{a^2+4a+4 + a^2-4a+4}{a^2-4} : \frac{16a}{a^2-4} = \frac{2a^2+8}{a^2-4} \cdot \frac{a^2-4}{16a} = \frac{2(a^2+4)}{16a} = \frac{a^2+4}{8a}$$