a) $\frac{4}{7}\cdot \left(\frac{1}{2}+1\frac{1}{3}+2\frac{1}{4}\right)=\frac{4}{7}\cdot \left(\frac{6}{12}+\frac{16}{12}+\frac{27}{12}\right)=\frac{4}{7}\cdot \frac{49}{12}=\frac{1}{1}\cdot \frac{7}{3}=\frac{7}{3}=2\frac{1}{3}$