Solution:

(i) $y(n)=x(n)+x(n-1)$

Given:

output of the system $y(n)=x(n)+x(n-1)$

If the input is delayed by ' $\mathrm{k}$ ' units in time, we have

$y(n, k)=x(n-k)+x(n-k-1)$

If the output is delayed by ' k ' units in time, then $(n-\gtn-k)$

$y(n-k)=x(n-k)+x(n-k-1)$

Here, $y(n, k)=y(n-k)$

Therefore, the system is time-invariant.

(ii) $y(n)=x(-n)$

Given:

output of the system $y(n)=x(-n)$

If the input is delayed by ' $\mathrm{k}$ ' units in time, we have,

$y(n, k)=x(-n-k)$

If the output is delayed by ' k ' units in time, then (n->n-k)

$y(n-k)=x(-n+k)$

Here, $y(n, k) \neq y(n-k)$

Therefore, the system is time-variant.