а) \(3\frac{2}{5}:\frac{17}{20}=\frac{17}{5}\cdot\frac{20}{17}=4\). Тогда \(\frac{11}{15}\cdot(4\frac{1}{2}-4)+1\frac{11}{20}=\frac{11}{15}\cdot\frac12+\frac{31}{20}=\frac{11}{30}+\frac{31}{20}=\frac{73}{60}=1\frac{13}{60}\).
б) \(5\frac47:1\frac5{21}=\frac{39}{7}:\frac{26}{21}=\frac{39}{7}\cdot\frac{21}{26}=\frac92\). Далее \(5\frac2{15}\cdot\frac3{22}=\frac{77}{15}\cdot\frac3{22}=\frac7{10}\), а \(1\frac{14}{15}=\frac{29}{15}\). Поэтому \(\frac92-\left(\frac7{10}+\frac{29}{15}\right)=\frac92-\frac{101}{30}=\frac{17}{15}=1\frac2{15}\).
Ответ: а) \(1\frac{13}{60}\); б) \(1\frac2{15}\).