1. \(\frac{41}{15}\cdot\left(4\frac{1}{2}-3\frac{2}{5}:\frac{17}{20}\right)+1\frac{11}{20}\)

\[3\frac{2}{5}:\frac{17}{20}=\frac{17}{5}\cdot\frac{20}{17}=4\]

\[4\frac{1}{2}-4=\frac{1}{2}\]

\[\frac{41}{15}\cdot\frac{1}{2}+1\frac{11}{20}=\frac{41}{30}+\frac{31}{20}\]

\[\frac{41}{30}+\frac{31}{20}=\frac{82}{60}+\frac{93}{60}=\frac{175}{60}=\frac{35}{12}=2\frac{11}{12}\]

Ответ: \(2\frac{11}{12}\).