$a)$
$Cu+2H_2SO_4\rightarrow CuSO_4+SO_2\uparrow+2H_2O$ $x 2x x$
$2Ag+2H_2SO_4\rightarrow Ag_2SO_4+SO_2\uparrow+2H_2O$
$y y 0,5y $
$\Rightarrow \begin{cases}64x+108y=34,4 \\2x+ y=\frac{32,61.1,84.98}{98.100} \end{cases}\Rightarrow \begin{cases}x=0,2 mol \\ y=0,2 mol \end{cases}$
$\Rightarrow \begin{cases}m_{Cu}=12,8 gam \\ m_{Ag}=21,6 gam\\V=22,4(x+0,5y)=6,72 lit \end{cases}$
$b)m_A=34,4+32,61.1,84-0,3.64=75,2 gam$
$\Rightarrow \begin{cases}\%m_{CuSO_4}=\frac{0,2.160.100}{75,2}=42,55 \% \\\%m_{Ag_2SO_4}=\frac{0,1.312.100}{75,2}=41,49\% \end{cases}$