To find the values of \( m \) such that \( A \cap B = \emptyset \), we first need to understand the intervals defined by \( A \) and \( B \).

### Given:
- Set \( A = \left[ m - 3, \frac{m + 2}{4} \right] \)
- Set \( B = (-\infty, -1) \cup [2, +\infty) \)

### Steps to solve:

1. **Find the conditions for \( A \cap B = \emptyset \)**:
- This means that the intervals in \( A \) must not overlap with any part of the intervals in \( B \).

2. **Find the bounds of set \( A \)**:
- The lower bound of \( A \) is \( m - 3 \).
- The upper bound of \( A \) is \( \frac{m + 2}{4} \).

3. **Identify the ranges for non-overlapping conditions**:
- For \( A \) not to overlap with \( B \):
1. The entire interval \( A \) must be less than -1, which leads to:
\[
\frac{m + 2}{4} < -1 \implies m + 2 < -4 \implies m < -6
\]
2. Alternatively, the entire interval \( A \) must be greater than or equal to 2:
\[
m - 3 \geq 2 \implies m \geq 5
\]

### Conclusion:
Thus, for \( A \cap B = \emptyset \), \( m \) should satisfy:
- \( m < -6 \) or \( m \geq 5 \).

The final solution is:
\[
m < -6 \quad \text{or} \quad m \geq 5
\]