a/ Gọi $n_{Zn}=x(mol);n_{Mg}=y(mol)$
$Zn+2HCl\to ZnCl_2+H_2\\\,\,x\quad\quad 2x\quad\quad\quad\quad\quad\quad\quad x$
$Mg+2HCl\to MgCl_2+H_2\\\,\,y\quad\quad\,\, 2y\quad\quad\quad\quad\quad\quad\quad y$
$\sum n_{H_2}=\dfrac{4,48}{22,4}=x+y=0,2(mol)$
$m_{hh}=65x+24y=8,9(g)$
$→$ Ta có hệ phương trình: $\begin{cases}x+y=0,2\\65x+24y=8,9\end{cases}$
$↔\begin{cases}x=0,1(mol)\\y=0,1(mol)\end{cases}$ hay $\begin{cases}n_{Zn}=0,1(mol)\\n_{Mg}=0,1(mol)\end{cases}$
$→\begin{cases}m_{Zn}=6,5g\\m_{Mg}=2,4 g\end{cases}$
Vậy $m_{Zn}=6,5g;\,m_{Mg}=2,4g$
b/ $n_{HCl}=2x+2y=2.0,1+2.0,1=0,4(mol)\\→m_{HCl}=0,4.36,5=14,6(g)\\→m_{dd\,HCl}=100(g)$
Vậy $m_{dd\,HCl}=100g$