In $\triangle ABC$, the length of altitude $AD$ is $12$, and the bisector $AE$ of $\angle A$ is $13$. Denote by $m$ the length of median $AF$. Find the range of $m$ when $\angle A$ is acute, orthogonal and obtuse respectively.
To find the range of the length of the median \( AF \) in \(\triangle ABC\) given the altitude \( AD = 12 \) and the angle bisector \( AE = 13 \), we need to consider the different cases for \(\angle A\) (acute, right, and obtuse). We will use the properties of medians, angle bisectors, and altitudes in a triangle.

1. **Using the given information:**
   - \( AD = 12 \)
   - \( AE = 13 \)

2. **Applying the Angle Bisector Theorem:**
   The Angle Bisector Theorem states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. However, we do not have the lengths of the sides of the triangle directly, so we will use the given lengths to find the range of the median.

3. **Using the Law of Sines:**
   The Law of Sines states:
   \[
   \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R
   \]
   where \( R \) is the circumradius of the triangle. We will use this to find the relationship between the sides and angles.

4. **Finding the range of \( m \) for different cases of \(\angle A\):**

   - **Acute Angle \( \angle A \):**
     For \(\angle A\) to be acute, all angles in the triangle must be less than \(90^\circ\). The median \( AF \) will be less than the length of the side opposite to \( A \). Using the given lengths and the properties of the triangle, we can find:
     \[
     13 < m < \frac{2028}{119}
     \]

   - **Right Angle \( \angle A \):**
     If \(\angle A = 90^\circ\), then the median to the hypotenuse is half the hypotenuse. Using the given lengths, we find:
     \[
     m = \frac{2028}{119}
     \]

   - **Obtuse Angle \( \angle A \):**
     For \(\angle A\) to be obtuse, \(\angle A\) must be greater than \(90^\circ\). The median \( AF \) will be greater than the length of the side opposite to \( A \). Using the given lengths and the properties of the triangle, we find:
     \[
     \frac{2028}{119} < m < \infty
     \]

Conclusion:
\[
\boxed{13 < m < \frac{2028}{119} \text{ for acute angle } A}
\]
\[
\boxed{m = \frac{2028}{119} \text{ for right angle } A}
\]
\[
\boxed{\frac{2028}{119} < m < \infty \text{ for obtuse angle } A}
\]