\(3\frac{5}{12}\cdot\left(4\frac{5}{6}-\frac{1}{2}+1\frac{1}{8}\right)\)

\(3\frac{5}{12}=\frac{41}{12}\), а \(4\frac{5}{6}-\frac{1}{2}+1\frac{1}{8}=\frac{29}{6}-\frac{1}{2}+\frac{9}{8}=\frac{131}{24}\).

\(\frac{41}{12}\cdot\frac{131}{24}=\frac{5371}{288}=18\frac{187}{288}\).

Ответ: \(18\frac{187}{288}\).