$\text{a)}$  Zauważmy, że:

$\frac{1}{10\cdot 11}=\frac{11-10}{10\cdot 11}=\frac{\sout{11}}{10\cdot \sout{11}}-\frac{\sout{10}}{\sout{10}\cdot 11}=\frac{1}{10}-\frac{1}{11}$  

Postępując analogicznie, otrzymamy:

$\frac{1}{11\cdot 12}=\frac{1}{11}-\frac{1}{12}$  

$\frac{1}{12\cdot 13}=\frac{1}{12}-\frac{1}{13}$  

i tak dalej...

Mamy więc:

$\frac{1}{10\cdot 11}+\frac{1}{11\cdot 12}+\frac{1}{12\cdot 13}+\dots +\frac{1}{88\cdot 89}+\frac{1}{89\cdot 90}=$  

$=\left(\frac{1}{10}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{12}\right)+\left(\frac{1}{12}-\frac{1}{13}\right)+\dots +\left(\frac{1}{88}-\frac{1}{89}\right)+\left(\frac{1}{89}-\frac{1}{90}\right)=$  

$=\frac{1}{10}-\cancel{\frac{1}{11}}+\cancel{\frac{1}{11}}-\cancel{\frac{1}{12}}+\cancel{\frac{1}{12}}-\cancel{\frac{1}{13}}+\dots +\cancel{\frac{1}{88}}-\cancel{\frac{1}{89}}+\cancel{\frac{1}{89}}-\frac{1}{90}=$  

$=\frac{1}{10}-\frac{1}{90}=\frac{9}{90}-\frac{1}{90}=\frac{8}{90}=\frac{4}{45}$