a)
\(n_{SO_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ m_{SO_2}=n\cdot M=0,2\cdot\left(32+16\cdot2\right)=12,8\left(g\right)\)
b)
\(n_{CH_4}=\dfrac{m}{M}=\dfrac{6,4}{12+1\cdot4}=0,4\left(mol\right)\\ V_{CH_4\left(dktc\right)}=n\cdot22,4=0,4\cdot22,4=8,96\left(l\right)\)