a)
$Na_2O + H_2O \to 2NaOH$
$n_{Na_2O} = \dfrac{6,2}{62} = 0,1(mol)$
$n_{NaOH} = 2n_{Na_2O} = 0,2(mol)$
$m_{NaOH} = 0,2.40 = 8(gam)$
b)
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{H_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,1(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,1.98}{20\% } = 49(gam)$
$V_{dd\ H_2SO_4} = \dfrac{49}{1,14} = 42,98(ml)$