$\text{a)}$

$1\frac{1}{5}\cdot 1\frac{2}{3}\cdot \frac{1}{7}+{\left(1\frac{1}{7}\right)}^2$ $=\frac{{\sout{6}}^2}{{\sout{5}}^1}\cdot \frac{{\sout{5}}^1}{{\sout{3}}^1}\cdot \frac{1}{7}+{\left(\frac{8}{7}\right)}^2$ $=\frac{2}{1}\cdot \frac{1}{7}+\frac{64}{49}=\frac{2}{7}+\frac{64}{49}=\frac{14}{49}+\frac{64}{49}$ $=\frac{78}{49}$ $=1\frac{29}{49}$