(i) Let the $3-digit$ number be $ABC,$ where $C$ is at the units place, $B$ at the tens place and $A$ at the hundreds place.

Now when repetition is allowed,

The number of digits possible at $C$ is $5.$ As repetition is allowed, the number of digits possible at $B$ and $A$ is also $5$ at each.

Hence, the total number possible $3-digit$ numbers

\[=5\text{ }\times \text{ }5\text{ }\times \text{ }5=125\]

(ii) Let the $3-digit$ number be $ABC,$ where $C$ is at the units place, $B$ at the tens place and $A$ at the hundreds place.

Now when repetition is not allowed,

The number of digits possible at $C$ is $5.$ Let’s suppose one of $5$ digits occupies place $C,$ now as the repletion is not allowed, the possible digits for place $B$ are $4$ and similarly there are only $3$ possible digits for place $A.$

Therefore, The total number of possible $3-digit$ numbers

\[=5\text{ }\times \text{ }4\text{ }\times \text{ }3=60\]