Let's solve question 16 step by step.

### a. Given \( A = \{ x \in \mathbb{R} : x \geq 2 \} \) and \( B = \{ x \in \mathbb{R} : x < 7 \} \)

1. \( A \cap B \) (intersection of A and B):
- \( A \cap B = \{ x : 2 \leq x < 7 \} \)

2. \( A \cup B \) (union of A and B):
- \( A \cup B = \{ x \in \mathbb{R} : x \geq 2 \text{ or } x < 7 \} \)
- Since \( x \geq 2 \) and \( x < 7 \) cover all real numbers, \( A \cup B = \mathbb{R} \)

3. \( A \setminus B \) (elements in A but not in B):
- \( A \setminus B = \{ x : x \geq 2 \text{ and } x \geq 7 \} \)
- Simplified to \( A \setminus B = \{ x : x \geq 7 \} \)

4. \( B \setminus A \) (elements in B but not in A):
- \( B \setminus A = \{ x : x < 7 \text{ and } x < 2 \} \)
- Simplified to \( B \setminus A = \{ x : x < 2 \} \)

5. \( \mathbb{R} \setminus A \) (complement of A in \( \mathbb{R} \)):
- \( \mathbb{R} \setminus A = \{ x : x < 2 \} \)

### b. Given \( A = \{ x \in \mathbb{R} : -1 < x \leq 2 \} \) and \( B = \{ x \in \mathbb{R} : 0 < x \leq 7 \} \)

1. \( A \cap B \):
- \( A \cap B = \{ x : 0 < x \leq 2 \} \)

2. \( A \cup B \):
- \( A \cup B = \{ x \in \mathbb{R} : -1 < x \leq 7 \} \)

3. \( A \setminus B \):
- \( A \setminus B = \{ x : -1 < x \leq 0 \} \)

4. \( B \setminus A \):
- \( B \setminus A = \{ x : 2 < x \leq 7 \} \)

5. \( \mathbb{R} \setminus A \):
- \( \mathbb{R} \setminus A = \{ x : x \leq -1 \text{ or } x > 2 \} \)

### c. Given \( A = \{ x \in \mathbb{R} : |x - 2| > 10 \} \) and \( B = \{ x \in \mathbb{R} : |x| < 3 \} \)

1. \( A \cap B \):
- From \( |x - 2| > 10 \), we have \( x < -8 \) or \( x > 12 \)
- From \( |x| < 3 \), we have \( -3 < x < 3 \)
- \( A \cap B = \varnothing \) (since no number can satisfy both conditions)

2. \( A \cup B \):
- The union is the combination of both ranges:
- \( A \cup B = \{ x : (-\infty, -8) \cup (-3, 3) \cup (12, \infty) \} \)

3. \( A \setminus B \):
- Elements in \( A \) but not in \( B \):
- \( A \setminus B = (-\infty, -8) \cup (12, \infty) \)

4. \( B \setminus A \):
- Elements in \( B \) but not in \( A \):
- \( B \setminus A = (-3, 3) \setminus ((-\infty, -8) \cup (12, \infty)) = (-3, 3) \)

5. \( \mathbb{R} \setminus A \):
- The complement of \( A \) in \( \mathbb{R} \):
- \( \mathbb{R} \setminus A = (-8, 12) \)

These are the solutions to the given sets for question 16.