a,
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$
$Mg+H_2SO_4\to MgSO_4+H_2$
$2Al+2NaOH+2H_2O\to 2NaAlO_2+3H_2$
b,
$n_{Mg}=\dfrac{0,6}{24}=0,025(mol)$
$n_{H_2}=\dfrac{1,568}{22,4}=0,07(mol)$
Theo PTHH:
$1,5n_{Al}+n_{Mg}=n_{H_2}$
$\Rightarrow n_{Al}=0,03$
$\%m_{Mg}=\dfrac{0,6.100}{0,6+0,03.27}=42,6\%$
$\Rightarrow \%m_{Al}=57,4\%$