Part A
Yes the system is time invariant.
$ x[n] \to (system) \to y[n] = (k+1)^2 x[n] \to (time shift) \to z[n] = y[n-(k+1)] = (k+1)^2 x[n-(k+1)] $
and then switch the system and time shift and compare to see if they are equal
$ x[n] \to (time shift) \to y[n] = x[n-(k+1)] \to (system) \to w[n] = (k+1)^2 y[n] = (k+1)^2 x[n-(k+1)] $
Since z[n] = w[n] the signal is time invariant.
