$a,\\ n_{HCl}=0,15.2=0,3(mol)\\ Đặt\ \begin{cases}n_{ZnO}=x(mol)\\n_{CuO}=y(mol)\end{cases}\\ ZnO+2HCl\to ZnCl_2+H_2O\\ CuO+2HCl\to CuCl_2+H_2O\\ \Rightarrow \begin{cases}81x+80y=12,1\\2x+2y=0,3\end{cases}\\ \Rightarrow \begin{cases}x=0,1\\y=0,05\end{cases}\\ \%m_{CuO}=\dfrac{0,05.80}{12,1}.100\%=33,06\%\\ \%m_{ZnO}=\dfrac{0,1.81}{12,1}.100\%=66,94\%\\ b,\\ ZnO+H_2SO_4\to ZnSO_4+H_2O\\ CuO+H_2SO_4\to CuSO_4+H_2O\\ n_{H_2SO_4}=n_{ZnO}+n_{CuO}=0,15(mol)\\ m_{dd\ H_2SO_4}=\dfrac{0,15.98}{20\%}=73,5(g)$

$c,\\ n_{ZnCl_2}=n_{ZnO}=0,1(mol)\\ n_{CuCl_2}=n_{CuO}=0,05(mol)\\ CM_{dd\ ZnCl_2}=\dfrac{0,1}{0,15}=\dfrac{2}{3}(M)\\ CM_{dd\ CuCl_2}=\dfrac{0,05}{0,15}=\dfrac{1}{3}(M)$