$PTHH :$
$NaOH+HCl\to NaCl+H_2O$
$m_{HCl}=300.36,5\%=10,95g$
$⇒n_{HCl}=\dfrac{10,95}{36,5}=0,3mol$
$Theo\ pt :$
$n_{NaOH}=n_{NaCl}=n_{HCl}=0,3mol$
$⇒m_{NaOH}=0,3.40=12g$
$m_{NaCl}=0,3.58,5=17,55g$
$m_{dd\ spu}=200+300=500g$
$⇒C\%_{NaOH}=\dfrac{12}{200}.100\%=6\%$
$C\%_{NaCl}=\dfrac{17.55}{500}=3,51\%$