a)$n_{H_2SO_4} = 0,3.1,5 = 0,45(mol)$
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} = 2n_{H_2SO_4} = 0,9(mol)$
$m_{dd\ NaOH} = \dfrac{0,9.40}{40\%} = 90(gam)$
b)
$n_{KOH} = n_{NaOH} = 0,9(mol0$
$m_{dd\ KOH} = \dfrac{0,9.56}{5,6\%} = 900(gam)$
$V_{dd\ KOH} = \dfrac{m}{D} = \dfrac{900}{1,045} = 861,24(ml)$