$Zn+2NaOH\rightarrow Na_2ZnO_2+H_2\uparrow$$x x$
$0,25 mol X:\begin{cases}n_{Zn}=n_{H_2} \\n_{Zn}+n_{Cu}=0,25 mol \end{cases}\Rightarrow \begin{cases}n_{Zn}=0,15 mol \\n_{Cu}=0,1 mol \end{cases}$
$\Rightarrow \begin{cases}\%m_{Zn}=\frac{0,15.65.100}{0,15.65+0,1.64}=60,27\% \\ \%m_{Cu}=39,63\% \end{cases}$