Помогите решить номер 4!

Решите задачу:

$a)\; \; (m-\frac{3}{7}):1\frac{1}{5}=3,5\\\\m-\frac{3}{7}=3,5\cdot \frac{6}{5}\\\\m-\frac{3}{7}=\frac{35}{10}\cdot \frac{6}{5}\\\\m-\frac{3}{7}=\frac{7\cdot 3}{5}\\\\m=\frac{21}{5}+\frac{3}{7}\\\\m=\frac{21\cdot 7+3\cdot 5}{35}\\\\m=\frac{162}{35}=4\frac{22}{35}\\\\b)\; \; (2,3-m):\frac{3}{5}=0,12\\\\2,3-m=0,12\cdot \frac{3}{5}\\\\2,3-m=\frac{12}{100}\cdot \frac{3}{5}\\\\2,3-m=\frac{9}{125}\\\\m=2,3-\frac{9}{125}=\frac{23}{10}-\frac{9}{125}=\frac{23\cdot 25-9\cdot 2}{250}=\frac{557}{250}=2,228$