$a)B=\dfrac{1}{4^2}+\dfrac{1}{5^2}+\cdots+\dfrac{1}{100^2}\\ B>\dfrac{1}{4.5}+\dfrac{1}{5.6}+\cdots+\dfrac{1}{100.101}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\cdots+\dfrac{1}{100}-\dfrac{1}{101}\\ =\dfrac{1}{4}-\dfrac{1}{101}\\ =\dfrac{97}{404}>\dfrac{1}{5}\\ B<\dfrac{1}{3.4}+\dfrac{1}{4.5}+\cdots+\dfrac{1}{99.100}\\ =\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\cdots+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{3}-\dfrac{1}{100}\\ =\dfrac{97}{300}>\dfrac{1}{3}\\ \Rightarrow \dfrac{1}{5}<B<\dfrac{1}{3}\\ b)C=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}+\cdots+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\\ 3C=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+\cdots+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\\ C+3C=1-\left(dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{3}{3^2}-\dfrac{2}{3^2}\right)-\left(\dfrac{4}{3^3}-\dfrac{3}{3^3}\right)+\cdots-\left(\dfrac{100}{3^{99}}-\dfrac{99}{3^{99}}\right)-\dfrac{100}{3^{100}}\\ \Leftrightarrow 4C=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+\cdots-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\\ 12C=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^3}-\cdots-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\\ 4C+12C=3-\dfrac{1}{3^{99}}-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}\\ \Leftrightarrow 16C=3-\dfrac{1}{3^{99}}-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}<3\\ \Rightarrow C<\dfrac{3}{16}$
