a,
$KOH+HCl\to KCl+H_2O$
b,
$m_{dd KOH}=50.11=550g$
$\Rightarrow n_{KOH}=\dfrac{550.20\%}{56}=\dfrac{55}{28}(mol)$
$\Rightarrow n_{HCl}=n_{KCl}=n_{KOH}=\dfrac{55}{28}(mol)$
$\Rightarrow m_{dd HCl}=36,5.\dfrac{55}{28}:10\%\approx 717g$
c,
$m_{KCl}=74,5.\dfrac{55}{28}\approx 146,34g$