Part A
It seems like this system that creates Y_k_[n]=(k+1)^2 d[n-(k+1)] from X_k_[n]=d[n-k] is not time invariant because at each k, the system creates a squared amplitude change in the graph of the original as the output. Because this ultimately changes the shape of the graph and the original is not preserved, this system is time variant.
Part B
Seeing as how the system shifts the delta function at k=0 by time 1, then to create the output y[n]=u[n-1] the input would only have to be x[n] = u[n] or the unit impulse.
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