$\text{a)}\left(2\frac{1}{3}+3\frac{1}{2}\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)=\left(\frac{7}{3}+\frac{7}{2}\right):\left(-\frac{25}{6}+\frac{22}{7}\right)=$  

$=\left(\frac{14}{6}+\frac{21}{6}\right):\left(-\frac{175}{42}+\frac{132}{42}\right)=\frac{35}{6}:\left(-\frac{43}{42}\right)=$  

$=\frac{35}{{\cancel{6}}_1}\cdot \left(-\frac{{\cancel{42}}^7}{43}\right)=-\frac{245}{43}$  

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$\text{b)}\left(\frac{7}{8}-5\frac{3}{16}\right)\cdot \frac{4}{3}+\frac{7}{10}:\left(\frac{2}{15}+\frac{4}{30}\right)=\left(\frac{14}{16}-5\frac{3}{16}\right)\cdot \frac{4}{3}+\frac{7}{10}:\left(\frac{2}{15}+\frac{2}{15}\right)=$  

$=\left(-4\frac{5}{16}\right)\cdot \frac{4}{3}+\frac{7}{10}:\frac{4}{15}=\left(-\frac{{\cancel{69}}^{23}}{{\cancel{16}}_4}\right)\cdot \frac{{\cancel{4}}^1}{{\cancel{3}}_1}+\frac{7}{{\cancel{10}}_2}\cdot \frac{{\cancel{15}}^3}{4}=$  

$=-\frac{23}{4}+\frac{21}{8}=-\frac{46}{8}+\frac{21}{8}=-\frac{25}{8}=-3\frac{1}{8}$