$2Al+6HCl\to 2AlCl_3+3H_2$
$AlCl_3+3NaOH\to Al(OH)_3+3NaCl$
a,
$m_{Ag}=10,8g$
$\Rightarrow m_{Al}=13,5-10,8=2,7g$
$\Rightarrow n_{Al}=\dfrac{2,7}{27}=0,1(mol)$
$n_{H_2}=\dfrac{3}{2}n_{Al}=0,15(mol)$
$\to V_2=0,15.22,4=3,36l$
b,
$n_{HCl}=2n_{H_2}=0,3(mol)$
$V_{HCl}=\dfrac{0,3}{1}=0,3l=300ml$
$\to V_1=300ml$
c,
$n_{Al(OH)_3}=n_{AlCl_3}=n_{Al}=0,1(mol)$
$\to a=m_{Al(OH)_3\downarrow}=0,1.78=7,8g$