$\begin{array}{l} a)\\Fe+H_2SO_4 \to FeSO_4+H_2\\n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\\n_{Fe}=n_{H_2}=0,4mol\\\%m_{Fe}=\dfrac{0,4.56}{28,8\%}=77,78\%\\\%m_{Cu}=100\%-77,78\%=22,22\%\\b)\\Fe+2HCl \to FeCl_2+H_2\\n_{Fe}=0,4mol\\m_{HCl}=300.14,6\%=43,8g\\n_{HCl}=\dfrac{43,8}{36,5}=1,2mol\\n_{Fe}=0,4mol<n_{HCl}=\dfrac{1,2}{2}=0,6mol\\n_{HCl(pu)}=2.0,4=0,8mol\\n_{HCl(du)}=1,2-0,8=0,4mol\\m_{HCl(du)}=0,4.36,5=14,6g\\mdd_{spu}=(0,4.56)+300-(0,4.2)=321,6g\\\%m_{HCl(du)}=\dfrac{14,6}{321,6\%}=4,54\%\\ m_{FeCl_2}=0,4.(56+35,5.2)=50,8g\\\%m_{FeCl_2}=\dfrac{50,8}{321,6\%}=15,8\%\end{array}$