Part E: Linearity and Time Invariance
Part A
For the output signal to be time invariant, the response to the shifted input signal $ x[n-N] $ should be the shifted output ($ y[n-N] $). Basically, this means that if the input signal is shifted along the x-axis by any amount of time, the output signal should produce the same value at $ n + N $ that it used to produce at $ n $ before the shift.
For the system given, let's use the signal/system corresponding to $ X $2 for the first one and $ X $3 for the second. Then select a time $ n $ of 4, and a shifted time $ N $ of 1. If this system was time invariant, $ Y $2$ [n] $ at time $ n = 5 $ should equal $ Y $1$ [n] $ at $ n = 4 $.
$ \,\ X $2$ \,\ [4] = \delta (2) $
$ \,\ Y $2$ \,\ [4] = 9\delta (1) $
$ \,\ X $3$ \,\ [5] = \delta (2) $
$ \,\ Y $3$ \,\ [5] = 16\delta (1) $
$ 9\delta (1) \ne 16\delta (1) $
$ \,\ Y $2$ \,\ [4] \ne Y $3$ \,\ [5] $
Therefore, the system is not time invariant.
Part B
We are given the premise that $ X[n] $ is linear. Because this is given to us, we can use the Time Invariance property which states that the output response of a shifted input signal is the shifted output response. That means all that would need to be done to produce an output of $ Y[n] = u[n-1] $ would be to use an input signal of $ X[n] = u[n] $.
