Time Invariance
For a given input $ X_i[n]=\delta[n-(i-1)] $ it yields the output $ Y_i[n]=(k+1)^2 \delta[n-(k+1)] $ is NOT time invariant.
Say we shifted the input at $ X_3[n] $ by +3, it would yield the output of $ Y[n]=\delta[n-1] $
However if we look at the output of $ X_3[n] $ and then shift it by 3, we get $ Y[n] = 16*\delta[n-4+3] = 16*\delta[n-1] $
Which is clearly 16 * the other input and thus they are not equal, so the system is not time invariant.
Linearity
Assuming the system is linear, $ \mu[n] $ should yield the output $ \mu[n-1] $ based on the response to $ \delta[n] $. Since the input of a function involving $ \delta[n] $ yields a time shifted delta function, a step function input should yield a time shifted step function for output.
