(2) In $\triangle A B C$, the altitude $A D=12$ from $A$ to side $B C$, and the angle bisector $A E=13$ of $\angle A$. Let the median $A F=m$ from $A$ to side $B C$. For what range of values of $m$ is $\angle A$ acute, right, or obtuse, respectively?
This is a plane geometry problem that is relatively close to textbook content. There are multiple methods to prove this problem. These methods can be roughly divided into two categories: one involves algebraic or trigonometric calculations, and the other focuses on pure geometric methods. Below, we provide two solutions.

Solution 1
First, by the Pythagorean theorem, we have \( E D = \sqrt{A E^2 - A D^2} = \sqrt{13^2 - 12^2} = \sqrt{(13-12)(13+12)} = 5 \). It is evident that regardless of how \(\triangle A B C\) changes, the right triangle \(\triangle A E D\) remains fixed. Let \(\alpha\) denote \(\angle E A D\), so \(\tan \alpha = 5 / 12\).
Let \(\beta = \angle B A E = \angle C A E\), hence
\[
\angle B A D = \beta + \alpha, \quad \angle C A D = \beta - \alpha
\]

Thus,
\[
\begin{array}{l}
B D = A D \tan (\beta + \alpha) = 12 \tan (\beta + \alpha) \\
C D = A D \tan (\beta - \alpha) = 12 \tan (\beta - \alpha)
\end{array}
\]

From \( B D = B F \pm F D \), \( C D = C F \mp F D \), and \( B F = C F \), we have
\[
\begin{array}{l}
F D = \frac{1}{2} |B D - C D| = 6 |\tan (\beta + \alpha) - \tan (\beta - \alpha)| = \\
6 \left| \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} - \frac{\tan \beta - \tan \alpha}{1 + \tan \beta \tan \alpha} \right| = \\
\left. 6 \left| \frac{2 \tan \alpha (1 + \tan^2 \beta)}{1 - \tan^2 \alpha \tan^2 \beta} \right| = 5 \frac{1 + \tan^2 \beta}{1 - \left( \frac{5}{12} \right)^2 \tan^2 \beta} \right| = \\
5 \frac{1 + \tan^2 \beta}{1 - \left( \frac{5}{12} \right)^2 \tan^2 \beta} \\
\end{array}
\]

We now consider three cases.
i. When \(\angle A\) is acute, \(0 < F G\); when \(\angle A\) is a right angle, \(A F = F G\); when \(\angle A\) is obtuse, \(A F < F G\).

Since \(A E\) is the angle bisector of \(\angle B A C\), the extension of \(A E\) must pass through point \(G\).
Thus,
\[
\frac{F G}{F E} = \frac{A D}{E D} = \frac{12}{5}
\]

This gives \(F G = \frac{12}{5} F E\). However,
\[
A F^2 = A D^2 + D F^2 = 12^2 + (F E + 5)^2
\]

So,
\[
\begin{array}{c}
F E = \sqrt{m^2 - 12^2} - 5 \\
F G = \frac{12}{5} \left( \sqrt{m^2 - 12^2} - 5 \right)
\end{array}
\]

Thus, the inequality \(F G < m\) is equivalent to
\[
\sqrt{m^2 - 12^2} - 5 < \frac{5}{12} m
\]

Removing the square root and simplifying, we get the inequality
\[
m^2 - \frac{600}{119} m - \frac{(12 \times 13)^2}{119} < 0
\]

Solving the above inequality, we get
\[
-12 < m < \frac{2028}{119}
\]

Since \(m\) cannot be negative, and by carefully examining the geometric figure, \(m = 13\) is impossible, otherwise it would lead to \(12 = 13\). Therefore, when \(\angle A\) is acute, it can only be \(13 < m < 2028 / 119\). For other cases, the results are as listed in Solution 1.