a,
$Na+HCl\to NaCl+\dfrac{1}{2}H_2$
$m_{Cu}=3,2g$
$\Rightarrow m_{Na}=5,5-3,2=2,3g$
$n_{Na}=\dfrac{2,3}{23}=0,1(mol)$
$\Rightarrow n_{H_2}=\dfrac{1}{2}n_{Na}=0,05(mol)$
$\to V_2=0,05.22,4=1,12l$
b,
$n_{HCl}=n_{Na}=0,1(mol)$
$\Rightarrow V_{HCl}=\dfrac{0,1}{1}=0,1l=100ml$
$\to V_1=100ml$
c,
$n_{NaCl}=n_{Na}=0,1(mol)$
$NaCl+AgNO_3\to AgCl+NaNO_3$
$\Rightarrow n_{AgCl}=n_{NaCl}=0,1(mol)$
$\to a=m_{AgCl}=0,1.143,5=14,35g$