$\begin{array}{l}a)\,\,\dfrac1{30}+\dfrac1{42}+\dfrac1{56}+\dfrac1{72}+\dfrac1{90}+\dfrac1{110}+\dfrac1{132}\\=\dfrac1{5.6}+\dfrac1{6.7}+\dfrac1{7.8}+\dfrac1{8.9}+\dfrac1{9.10}+\dfrac1{10.11}+\dfrac1{11.12}\\=\dfrac15-\dfrac16+\dfrac16-\dfrac17+\dfrac17-\dfrac18\,+\,...+\,\dfrac1{11}-\dfrac1{12}\\=\dfrac15-\dfrac1{12}\\=\dfrac7{60}\\\,\\b)\,\,\dfrac1{1.2.3}+\dfrac1{2.3.4}+\dfrac1{3.4.5}\,+\,...+\,\dfrac1{10.11.12}\\=\dfrac12\left(\dfrac2{1.2.3}+\dfrac2{2.3.4}+\dfrac2{3.4.5}\,+\,...+\,\dfrac2{10.11.12}\right)\\=\dfrac12\left(\dfrac1{1.2}-\dfrac1{2.3}+\dfrac1{2.3}-\dfrac1{3.4}+\dfrac1{3.4}-\dfrac1{4.5}\,+\,...+\,\dfrac1{10.11}-\dfrac1{11.12}\right)\\=\dfrac12\left(\dfrac1{1.2}-\dfrac1{11.12}\right)\\=\dfrac12\left(\dfrac12-\dfrac1{132}\right)\\=\dfrac12\cdot\dfrac{65}{132}\\=\dfrac{65}{264} \end{array}$