$n_{HCl}=\dfrac{100.3,65\%}{36,5}=0,1(mol)$
$n_{NaOH}=\dfrac{200.4\%}{40}=0,2(mol)$
$NaOH+HCl\to NaCl+H_2O$
$\Rightarrow NaOH$ dư
$m_{dd}=100+200=300g$
$n_{NaCl}=n_{HCl}=0,1(mol)\Rightarrow C\%_{NaCl}=\dfrac{0,1.58,5.100}{300}=1,95\%$
$n_{NaOH\text{dư}}=0,2-0,1=0,1(mol)\Rightarrow C\%_{NaOH}=\dfrac{0,1.40.100}{300}=1,33\%$