$n_{HCl}=\dfrac{0,56}{22,4}=0,025(mol)$
$m_{dd AgNO_3}=50.1,1=55g$
$\to n_{AgNO_3}=\dfrac{55.8\%}{170}=0,026(mol)$
$AgNO_3+HCl\to AgCl+HNO_3$
$\to n_{AgCl\downarrow}=n_{HNO_3}=0,025(mol)$
$m_{dd \text{spu}}=55+0,025.36,5-0025.143,5=52,325g$
$\to C\%_{HNO_3}=\dfrac{0,025.63.100}{52,325}=3\%$