$a)1\frac{1}{3}-\frac{1}{{\sout{3}}^1}\cdot \frac{{\sout{6}}^2}{7}\cdot \frac{14}{15}\cdot \frac{5}{4}=1\frac{1}{3}-\frac{2}{{\sout{7}}^1}\cdot \frac{{\sout{14}}^2}{15}\cdot \frac{5}{4}=1\frac{1}{3}-\frac{4}{{\sout{15}}^3}\cdot \frac{{\sout{5}}^1}{4}=1\frac{1}{3}-\frac{1}{3}=1$

$b)12\frac{5}{7}-1\frac{2}{7}+2\frac{3}{7}=11\frac{3}{7}+2\frac{3}{7}=13\frac{6}{7}$