a, 
$n_{H_2SO_4}=0,2 mol$
$2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$
$\Rightarrow n_{NaOH}=0,2.2=0,4 mol$
$\to m_{\text{dd NaOH}}=0,4.40:20\%=80g$
b,
$2KOH+H_2SO_4\to K_2SO_4+2H_2O$
$\Rightarrow n_{KOH}=0,2.2=0,4 mol$
$m_{\text{dd KOH}}=0,4.56:5,6\%=400g$
$\to V_{\text{dd KOH}}=\dfrac{400}{1,045}=382,775(ml)$