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The problem asks to solve the following sets by extension (that is, to list the elements explicitly):

1. \( E = \{ x \in \mathbb{N} \mid x^2 = 16 \} \)
2. \( F = \{ x \in \mathbb{N} \mid x - 3 = 2 \} \)

Let's break them down:

### 1. Set \( E \)
We are given the equation \( x^2 = 16 \), and \( x \in \mathbb{N} \) (the natural numbers). The natural numbers are positive integers, so we solve:
\[
x^2 = 16 \quad \Rightarrow \quad x = 4 \quad \text{(since \( x \) must be positive)}
\]
Thus, the set \( E \) by extension is:
\[
E = \{ 4 \}
\]

### 2. Set \( F \)
We are given the equation \( x - 3 = 2 \), and \( x \in \mathbb{N} \). Solving for \( x \):
\[
x - 3 = 2 \quad \Rightarrow \quad x = 5
\]
Thus, the set \( F \) by extension is:
\[
F = \{ 5 \}
\]

### Final Answer:
- \( E = \{ 4 \} \)
- \( F = \{ 5 \} \)

Would you like further details or have any questions?

Here are some related questions to explore further:
1. What if we expanded the domain from natural numbers to integers?
2. How would the solution change if the equation for \( E \) was \( x^2 = 25 \)?
3. What happens if the condition for \( F \) were \( x - 3 \leq 2 \)?
4. Can you explain the concept of solving sets by extension vs. by comprehension?
5. How do you handle cases where there are no solutions within \( \mathbb{N} \)?

**Tip**: When solving for natural numbers, always remember that \( \mathbb{N} \) typically includes only positive integers, though in some contexts it includes zero as well!

Resolver por extensión los siguientes conjuntos: 1. E = {x ∈ N / x^2 = 16}, 2. F = {x ∈ N / x - 3 = 2}

Learn how to solve sets defined by quadratic and linear equations using natural numbers. This problem involves solving x^2 = 16 and x - 3 = 2, suitable for students in Grades 6-8.

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