1. In the quadratic equation $x^{2}+z_{1} x+z_{2}+m=0$, $z_{1}, z_{2}, m$ are all complex numbers, and $z_{1}^{2}-4 z_{2}=16+20 \mathrm{i}$. Suppose the two roots $\alpha, \beta$ of this equation satisfy $|\alpha-\beta|=2 \sqrt{7}$, find the maximum and minimum values of $|m|$.
$$
\begin{array}{l}
(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=z_{1}^{2}-4\left(z_{2}+m\right)=16+20 \mathrm{i}-4 m \\
\Rightarrow|16+20 \mathrm{i}-4 m|=\left|(\alpha-\beta)^{2}\right|=|\alpha-\beta|^{2}=28 .
\end{array}
$$

Since $|16+20 \mathrm{i}-4 m| \leqslant|16+20 \mathrm{i}|+4|m| \Rightarrow 4|m| \geqslant 28-|16+20 \mathrm{i}|=28-4 \sqrt{41}$ $\Rightarrow|m| \geqslant 7-\sqrt{41}$, equality holds when $m=\frac{\sqrt{41}-7}{\sqrt{41}}(4+5 \mathrm{i})$;
$$
|16+20 \mathrm{i}-4 m| \geqslant 4|m|-|16+20 \mathrm{i}| \Rightarrow 4|m| \leqslant 28+|16+20 \mathrm{i}|=28+4 \sqrt{41}
$$
$\Rightarrow|m| \leqslant 7+\sqrt{41}$, equality holds when $m=\frac{7+\sqrt{41}}{\sqrt{41}}(4+5 \mathrm{i})$.
In summary, the maximum value of $|m|$ is $7+\sqrt{41}$, and the minimum value is $7-\sqrt{41}$.