Q.Write the set $A = \{x : x \in \mathbb{N},\ x \text{ is a divisor of } 12\}$ in roster form, and write the set $\{2, 4, 6, 8, 10\}$ in set-builder form.

Converting between the two forms just means either listing the elements that satisfy the stated rule, or spotting the common property of a listed set.

The natural divisors of $12$ are $1, 2, 3, 4, 6, 12$, so $A = \{1, 2, 3, 4, 6, 12\}$. The set $\{2,4,6,8,10\}$ is the even natural numbers not exceeding $10$.

Roster form of $A$. A natural number is a divisor of $12$ if it divides $12$ exactly (no remainder). Testing the naturals: $1, 2, 3, 4, 6$ and $12$ all divide $12$, while $5, 7, 8, 9, 10, 11$ do not. Listing the divisors:

Set-builder form of $\{2, 4, 6, 8, 10\}$. These are precisely the even natural numbers from $2$ up to $10$. The common property "is an even natural number and is at most $10$" describes exactly this list, so

(An equivalent set-builder description is $\{2n : n \in \mathbb{N},\ 1 \le n \le 5\}$.)

For the set-builder answer, the generator form $\{2n : n \in \mathbb{N},\ 1 \le n \le 5\}$ is equally correct and is often preferred when the pattern (a multiple) is the point.

Common slips: forgetting that $1$ and the number $12$ itself both count as divisors, or omitting a divisor such as $6$. In set-builder form, dropping the domain condition $x \in \mathbb{N}$ would wrongly admit numbers like $2.5$ or negative evens.