One, (This question is worth 25 points) In the quadratic equation of $x$, $x^{2}+z_{1} x+z_{2}+m=0$, $z_{1}, z_{2}, m$ are all complex numbers, and $z_{1}^{2}-4 z_{2}=$ $16+20 i$. Suppose the two roots $\alpha, \beta$ of this equation satisfy $|\alpha-\beta|=2 \sqrt{7}$, find the maximum and minimum values of $|m|$.
$$
\sqrt{41}+7,7-\sqrt{41}
$$

I. 【Analysis and Solution】According to the expression theorem, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
\alpha+\beta=-z_{1} \\
\alpha \beta=z_{2}+m
\end{array}\right. \\
\because(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=z_{1}^{2}-4 z_{2}-4 m, \\
\therefore|\alpha-\beta|^{2}=\left|4 m-\left(z_{1}^{2}-4 z_{2}\right)\right|=28, \\
\therefore\left|m-\frac{1}{4}\right|\left(z_{1}^{2}-4 z_{2}\right)=7 .
\end{array}
$$

This means $|m-4+5 i|=7$.
This indicates that the complex number $m$ lies on the circumference of a circle with center $A(4,5)$ and radius 7. Since $O$ is inside the circle $\odot A$ with $|O A|=$ $\sqrt{4^{4}+5^{2}}=\sqrt{41}<7$, the origin $O$ is inside $\odot A$. Connecting $O A$ and extending it to intersect $\odot A$ at points $B$ and $C$, we have $|O B|=\mid O A$ $|+| A B \mid=\sqrt{41}+7$ as the maximum value of $|m|$, and $|O C|=|C A|-|A O|=7-\sqrt{41}$ as the minimum value of $|m|$.
$\therefore|m|$ has a maximum value of $\sqrt{41}+7$ and a minimum value of $7-\sqrt{41}$.