Выполним задание:
$$√{a^2+8ab+16b^2} = √{(3\frac{2}{3})^2 + 8 \cdot 3\frac{2}{3} \cdot \frac{1}{3} + 16 \cdot (\frac{1}{3})^2}$$
$$3\frac{2}{3} = \frac{3 \cdot 3 + 2}{3} = \frac{9 + 2}{3} = \frac{11}{3}$$
$$√{(\frac{11}{3})^2 + 8 \cdot \frac{11}{3} \cdot \frac{1}{3} + 16 \cdot (\frac{1}{3})^2} = √{\frac{121}{9} + \frac{88}{9} + \frac{16}{9}} = √{\frac{121 + 88 + 16}{9}} = √{\frac{225}{9}} = \frac{√{225}}{√{9}} = \frac{15}{3} = 5$$
Ответ: 5