Part A
This system is not time invariant because it depends on the time shift of the function. This System goes in the order of:
X1(t) --> delay --> System --> multiplication and addition --> Y(t)
However, if it goes in this order:
X1(t) --> multiplication and addition --> system --> delay --> Y(t)
Then the two outputs would not be equal. The coefficients would be different.
Part B
With only inputing
x[n] = u[n]
Then the coefficient of the output is just the square of the time time delay value, which this time delay would be "1".
So u^2 is just (1)^2 = 1
So the output of the signal is:
y[n] = (u * 1)[n - 1] = u[n - 1]