\(a.\)
\(n_{KClO_3}=\dfrac{3.675}{122.5}=0.03\left(mol\right)\)
\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
\(0.03........................0.045\)
\(V_{O_2}=0.045\cdot22.4=1.008\left(l\right)\)
\(n_{O_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)
\(\Rightarrow n_{KClO_3}=\dfrac{0.5\cdot2}{3}=\dfrac{1}{3}mol\)
\(\Rightarrow m_{KClO_3}=\dfrac{1}{3}\cdot122.5=40.83\left(g\right)\)