Решение

1. a) $$\sqrt[4]{8} \cdot \sqrt{2} = \sqrt[4]{2^{3}} \cdot \sqrt{2} = 2^{\frac{3}{4}} \cdot 2^{\frac{1}{2}} = 2^{\frac{3}{4} + \frac{1}{2}} = 2^{\frac{3}{4} + \frac{2}{4}} = 2^{\frac{5}{4}} = \sqrt[4]{2^{5}} = \sqrt[4]{32}$$
2. б) $$\sqrt[6]{2^{11} \cdot 2^{7} \cdot 3^{12}} = \sqrt[6]{2^{18} \cdot 3^{12}} = 2^{\frac{18}{6}} \cdot 3^{\frac{12}{6}} = 2^{3} \cdot 3^{2} = 8 \cdot 9 = 72$$

Ответ: a) $$\sqrt[4]{32}$$; б) 72