5210 $\star \star$ For the equation $x^{2}+z_{1} x+z_{2}+m=0$ in terms of $x$, where $z_{1}, z_{2}, m$ are complex numbers, and $z_{1}^{2}-4 z_{2}=16+20 \mathrm{i}$. Let the two roots of this equation be $\alpha, \beta$, satisfying $|\alpha-\beta|=2 \sqrt{7}$, find the maximum and minimum values of $|m|$.
Solving using Vieta's formulas, we get 
\[
\left\{\begin{array}{l}
\alpha+\beta=-z_{1}, \\
\alpha \beta=z_{2}+m .
\end{array}\right.
\]
Thus, 
\[
(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=z_{1}^{2}-4 z_{2}-4 m,
\]
so 
\[
|\alpha-\beta|^{2}=\left|4 m-\left(z_{1}^{2}-4 z_{2}\right)\right|=28,
\]
hence 
\[
|m-(4+5i)|=7.
\]

This indicates that the point corresponding to the complex number \( m \) lies on the circumference of a circle with center \( A(4,5) \) and radius 7, and the origin \( O \) is inside the circle \( \odot A \). As shown in the figure. Connecting \( O A \) and extending it to intersect \( \odot A \) at points \( B \) and \( C \). Thus, we can easily find
\[
\begin{array}{l}
|m|_{\min }=|O C|-|C A|-|O A|=7-\sqrt{41}, \\
|m|_{\max }=|O B|-|O A|+|A B|=7+\sqrt{41} .
\end{array}
\]