2) \(1\frac{7}{48}\cdot2\frac{2}{5}-\left(9\frac{1}{6}\cdot\frac{4}{15}+2\frac{5}{9}\right)\cdot\frac{2}{5}\);

Преобразуем смешанные числа:

\(1\frac{7}{48}=\frac{55}{48},\quad 2\frac{2}{5}=\frac{12}{5},\quad 9\frac{1}{6}=\frac{55}{6},\quad 2\frac{5}{9}=\frac{23}{9}.\)

\[\frac{55}{48}\cdot\frac{12}{5}=\frac{11}{4}.\]

В скобках:

\[\frac{55}{6}\cdot\frac{4}{15}+\frac{23}{9}=\frac{22}{9}+\frac{23}{9}=5.\]

Тогда

\[\frac{11}{4}-5\cdot\frac{2}{5}=\frac{11}{4}-2=\frac{3}{4}.\]

Ответ: \(\frac{3}{4}\).