a) $$\frac{81^6}{16^{12}} = \frac{(3^4)^6}{(2^4)^{12}} = \frac{3^{24}}{2^{48}} = \frac{3^{24}}{(2^2)^{24}} = \frac{3^{24}}{4^{24}} = (\frac{3}{4})^{24}$$
б) $$\frac{81^{25}}{27^{33}} = \frac{(3^4)^{25}}{(3^3)^{33}} = \frac{3^{100}}{3^{99}} = 3^{100-99} = 3^1 = 3$$