Discrete time Signal Transformation
The input to the system $x_1[n]=\delta[n-5]$ is zero everywhere and is only one when the argument of delta is zero (at $n=5$). The output $$y_1[n]=x_1[5n+2]=\delta[(5n+2)-5]$$ would be zero everywhere too, except when the argument of $\delta$ is zero: $$5n+2-5=0$$ which requires $n=\frac{3}{5}$. But since $n$ only has integer values, it can never be achieved. Hence, the output of the system is zero always.