6 a) The system cannot be time-invariant.
$ X_k[n] = \delta[n - k] \rightarrow system \rightarrow Y_k[n] = (k + 1)^2 \delta[n - (k + 1)] $
Let us apply a time-delay of $ n_0 $ to the system.
System followed by time-delay:
$ \delta[n - k] \rightarrow system \rightarrow (k + 1)^2 \delta[n - (k + 1)] \rightarrow time-delay \rightarrow (k + 1)^2 \delta[n - n_0 -(k + 1)] = (k + 1)^2 \delta[n -(k + 1 +n_0)] $
Time-delay followed by system:
$ \delta[n - k] \rightarrow time-delay \rightarrow \delta[n-(k + n_0)] \rightarrow system \rightarrow (k + n_0 + 1)^2 \delta[n - (k + n_0 + 1)] $
Since the outputs don't match, the system is not time-invariant.
6 b) Since the system is linear, the input needed to yield Y[n] = u[n - 1] is X[n] = u[n].
$ X_k[n] = u[n - k] \rightarrow system \rightarrow Y_k[n] = (k + 1)^2 u[n - (k + 1)] $
In this case k = 0.
Therefore, $ X_0[n] = u[n] \rightarrow system \rightarrow Y_0[n] = u[n + 1)] $
