a,
$n_{CaO}=\dfrac{14}{56}=0,25(mol)$
$CaO+H_2O\to Ca(OH)_2$
$\to n_{Ca(OH)_2}=n_{CaO}=0,25(mol)$
$n_{CO_2}=\dfrac{4,48}{22,4}=0,2(mol)$
$CO_2+Ca(OH)_2\to CaCO_3+H_2O$
$\dfrac{0,25}{1}>\dfrac{0,2}{1}\to CO_2$ hết, $Ca(OH)_2$ dư
$\to n_{CaCO_3}=n_{CO_2}=0,2(mol)$
$\to m=m_{CaCO_3}=0,2.100=20g$
b,
$n_{Ca(OH)_2\text{dư}}=0,25-0,2=0,05(mol)$
$\to m_{Ca(OH)_2}=0,05.74=3,7g$