$\text{a)}\sqrt[3]{8\cdot 27}=\sqrt[3]{8}\cdot \sqrt[3]{27}=\sqrt[3]{2^3}\cdot \sqrt[3]{3^3}=2\cdot 3=6$

$\text{b)}\sqrt[3]{5}\cdot \sqrt[3]{25}=\sqrt[3]{5\cdot 25}=\sqrt[3]{5^3}=5$

$\text{c)}\sqrt[3]{28}\cdot \sqrt[3]{98}=\sqrt[3]{28\cdot 98}=\sqrt[3]{4\cdot 7\cdot 49\cdot 2}=\sqrt[3]{2^3\cdot 7^3}=\sqrt[3]{2^3}\cdot \sqrt[3]{7^3}=2\cdot 7=14$  

$\text{d)}\frac{\sqrt[3]{56}}{\sqrt[3]{7}}=\sqrt[3]{\frac{56}{7}}=\sqrt[3]{8}=\sqrt[3]{2^3}=2$  

$\text{e)}\sqrt[3]{\frac{27}{64}}-\sqrt[3]{\frac{125}{216}}=\sqrt[3]{\frac{3^3}{4^3}}-\sqrt[3]{\frac{5^3}{6^3}}=\sqrt[3]{{\left(\frac{3}{4}\right)}^3}-\sqrt[3]{{\left(\frac{5}{6}\right)}^3}=\frac{3}{4}-\frac{5}{6}=\frac{9}{12}-\frac{10}{12}=-\frac{1}{12}$

$\text{f)}\frac{\sqrt[3]{40}}{\sqrt[3]{5}}+\frac{\sqrt[3]{3}}{\sqrt[3]{81}}=\sqrt[3]{\frac{40}{5}}+\sqrt[3]{\frac{3}{81}}=\sqrt[3]{8}+\sqrt[3]{\frac{1}{27}}=\sqrt[3]{2^3}+\sqrt[3]{{\left(\frac{1}{3}\right)}^3}=2+\frac{1}{3}=2\frac{1}{3}$