Let the function $f(x)=x^{2}+a x+b(a, b \in$
$\mathbf{R})$. If there exists a real number $m$, such that
$$
|f(m)| \leqslant \frac{1}{4} \text {, and }|f(m+1)| \leqslant \frac{1}{4},
$$

find the maximum and minimum values of $\Delta=a^{2}-4 b$.
Solution 1: If $\Delta=a^{2}-4 b$
$$
\frac{-a-\sqrt{\Delta+1}}{2} \leqslant x \leqslant \frac{-a-\sqrt{\Delta-1}}{2}
$$

or $\frac{-a+\sqrt{\Delta-1}}{2} \leqslant x \leqslant \frac{-a+\sqrt{\Delta+1}}{2}$.
If $|f(m)| \leqslant \frac{1}{4}$, and $|f(m+1)| \leqslant \frac{1}{4}$, then it must be true that
$$
\frac{-a+\sqrt{\Delta-1}}{2}-\frac{-a-\sqrt{\Delta-1}}{2} \leqslant 1
$$

or $\frac{-a-\sqrt{\Delta-1}}{2}-\frac{-a-\sqrt{\Delta+1}}{2} \geqslant 1$
or $\frac{-a+\sqrt{\Delta+1}}{2}-\frac{-a+\sqrt{\Delta-1}}{2} \geqslant 1$,
which means $\sqrt{\Delta-1} \leqslant 1$ or $\sqrt{\Delta+1}-\sqrt{\Delta-1} \geqslant 1$.
Solving this, we get $\Delta \leqslant 2$.
When $a=1, b=-\frac{1}{4}$,
$|f(0)|=|f(-1)|=\frac{1}{4}$, and $\Delta=2$,
Therefore, the maximum value of $\Delta$ is 2.
Solution 2: Since $1=\left|\left(m+1-x_{0}\right)-\left(m-x_{0}\right)\right|$
$$
\leqslant\left|m+1-x_{0}\right|+\left|m-x_{0}\right|,
$$

thus, $\left|m+1-x_{0}\right|$ and $\left|m-x_{0}\right|$ must have one that is not less than $\frac{1}{2}$.
If $\Delta=a^{2}-4 b<0$, then
$$
\begin{array}{l}
f(m)=\left(m+\frac{a}{2}\right)^{2}-\frac{\Delta}{4}, \\
f(m+1)=\left(m+1+\frac{a}{2}\right)^{2}-\frac{\Delta}{4}
\end{array}
$$

one of them must be greater than $\frac{1}{4}$.
This contradicts the given condition, so, $\Delta \geqslant 0$.
The rest is the same as Solution 1.