Case 1: Task-2 is assigned to Person-3

Task-1 Assignment: Since Task-1 cannot be assigned to Person-1 or Person-2, it must be assigned to one of the remaining persons: Person-4 or Person-5. This gives us 2 choices.

Remaining Tasks: After assigning Tasks-1 and -2, we have 3 tasks left (Tasks-3, -4, and -5) and 3 persons left (Persons-1, -2, and -5). These remaining tasks can be assigned to the remaining persons in any order, which can be done in $3!$ (3 factorial) ways.

Therefore, the total number of assignments for Case 1 is:

2 (choices for Task-1) × $3!$ (ways to assign remaining tasks) = 2 × 6 = 12 ways.

Case 2: Task-2 is assigned to Person-4

Task-1 Assignment: Similarly, Task-1 must be assigned to one of the remaining persons: Person-3 or Person-5. This gives us 2 choices.

Remaining Tasks: After assigning Tasks-1 and -2, we have 3 tasks left (Tasks-3, -4, and -5) and 3 persons left (Persons-1, -2, and -3). These remaining tasks can be assigned to the remaining persons in any order, which can be done in $3!$ ways.

Therefore, the total number of assignments for Case 2 is:

2 (choices for Task-1) × $3!$ (ways to assign remaining tasks) = 2 × 6 = 12 ways.

Total Number of Assignments:

Adding the results from both cases, we get:

12 (from Case 1) + 12 (from Case 2) = 24 ways.

Therefore, the assignment can be done in 24 ways.

The correct answer is C. 24.