Part (a)
No. This system is not time-invariant. The general equation of the system is as follows.
$ X_{k}[n] = d[n-k] $
$ Y_{k}[n] = (k+1)^2 d[n-(k+1)] $
Shifting $ X_{k}[n] $ by a constant "a" yields $ X_{k}[n-a] $
Shifting $ Y_{k}[n] $ by a constant "a" yields $ Y_{k}[n-a] $
$ (1) X_{k}[n-a] = d[n-k-a] $
$ (2) Y_{k}[n-a] = (k+1)^2 d[n-(k+1)-a] $
As shown in (2), $ (k+1)^2 $ does not get shifted. Thus, a shift made in $ X_{k}[n] $ does not accordingly shift $ Y_{k}[n] $
Part (b)
$ X[n] = u[n] $
This would yield the expected result.
Since the system is linear, the input u[n] would result in u[n-1] as d[n] resulted in d[n-1]
