Twenty tickets are numbered 1 to 20. Let A be the event "the number is a multiple of 3", and let B…

Step 1. The counts. $A = \{3, 6, 9, 12, 15, 18\}$, so $n(A) = 6$. $B = \{4, 8, 12, 16, 20\}$, so $n(B) = 5$. Only $12$ is in both, so $n(A \cap B) = 1$.

Step 2. The union. $n(A \cup B) = 6 + 5 - 1 = 10$.

Step 3. Neither. $n((A \cup B)') = 20 - 10 = 10$. Now count $A' \cap B'$ directly: the tickets that are not multiples of $3$ and not multiples of $4$ are $10$ in number. De Morgan's law $(A \cup B)' = A' \cap B'$ holds.

Step 4. The complement of the overlap. $n((A \cap B)') = 20 - 1 = 19$. This equals $n(A' \cup B')$, which is the second law.