3\frac{1}{5}-1\frac{3}{10}+(4\frac{2}{9}-2\frac{1}{6}+\frac{5}{12})\cdot\frac{7}{15}\cdot\frac{3}{8}

1. $$3\frac{1}{5}-1\frac{3}{10} = \frac{16}{5} - \frac{13}{10} = \frac{32}{10} - \frac{13}{10} = \frac{19}{10}$$

2. $$4\frac{2}{9}-2\frac{1}{6}+\frac{5}{12} = \frac{38}{9} - \frac{13}{6} + \frac{5}{12} = \frac{38 \cdot 4}{9 \cdot 4} - \frac{13 \cdot 6}{6 \cdot 6} + \frac{5 \cdot 3}{12 \cdot 3} = \frac{152}{36} - \frac{78}{36} + \frac{15}{36} = \frac{152 - 78 + 15}{36} = \frac{89}{36}$$

3. $$\frac{89}{36} \cdot \frac{7}{15} \cdot \frac{3}{8} = \frac{89}{36} \cdot \frac{7 \cdot 3}{15 \cdot 8} = \frac{89}{36} \cdot \frac{21}{120} = \frac{89 \cdot 7}{12 \cdot 40} = \frac{623}{480}$$

4. $$\frac{19}{10} + \frac{623}{480} = \frac{19 \cdot 48}{10 \cdot 48} + \frac{623}{480} = \frac{912}{480} + \frac{623}{480} = \frac{912 + 623}{480} = \frac{1535}{480} = \frac{307}{96} = 3\frac{19}{96}$$

Ответ: $$3\frac{19}{96}$$