E-a
Yes,the system can be time-invariant.
the system is $ Y_k[n] = (k + 1)^2 *X_k[n-1] $
$ X_k[n]\rightarrow system \rightarrow Y_k[n] = (k + 1)^2*X_k[n-1]\rightarrow Time Delay by m\rightarrow Z_k[n]=(k+1)^2*X_k[n-m-1] $
$ X_k[n]\rightarrow Time Delay by m\rightarrow Y_k[n] = X_k[n-m]\rightarrow Time System\rightarrow Z_k[n]=Y_k[n-m]=(k+1)^2*X_k[n-m-1] $
Since the outputs match, the system is time-invariant.
E-b
Since the system is linear, the input should be X[n] = u[n] in order to get Y[n]=u[n-1], where k=0
