Part A
If we use the same procedure as we did in HW2D then we will find out it is not time variant. The we get
$ \,\ y(t)= (k+1)^2 \delta[(n-no)-(k + 1)] $
$ \,\ z(t) = (k+1+no)^2 \delta[(n-no)-(k + 1)] $
Part B
If we set k = 0 then the answer seems easier to get
i.e. $ \,\ \delta(n - 1) $ and now $ \,\ X[n] = u(n) $
