a,
$Mg+2HCl\to MgCl_2+H_2\ \ (1)$
$Al_2O_3+6HCl\to 2AlCl_3+3H_2O\ \ (2)$
$n_{H_2}=\dfrac{4,48}{22,4}=0,2\ (mol)$
$\to n_{Mg}=n_{H_2}=0,2\ (mol)$
$\to m_{Mg}=0,2.24=4,8\ (g)$
$\to m_{Al_2O_3}=15-4,8=10,2\ (g)$
b,
$n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\ (mol)$
$\to n_{HCl\ (2)}=6n_{Al_2O_3}=6.0,1=0,6\ (mol)$
$n_{HCl\ (1)}=2n_{H_2}=2.0,2=0,4\ (mol)$
$\to \sum n_{HCl}=0,4+0,6=1\ (mol)$
$\to V_{ddHCl}=\dfrac{1}{2}=0,5\ (l)=500\ (ml)$