One, (20 points) Given the equation $a x^{2}+4 x+b=0$ $(a<0)$ has two real roots $x_{1} 、 x_{2}$, and the equation $a x^{2}+3 x+b=0$ has two real roots $\alpha 、 \beta$.
(1) If $a 、 b$ are both negative integers, and $|\alpha-\beta|=1$, find the values of $a 、 b$;
(2) If $\alpha<1<\beta<2, x_{1}<x_{2}$, prove:
$-2<x_{1}<1<x_{2}$.
(1) From the problem, we have $\alpha+\beta=-\frac{3}{a}, \alpha \beta=\frac{b}{a}$.
From $|\alpha-\beta|=1 \Rightarrow(\alpha+\beta)^{2}-4 \alpha \beta=1$
$$
\Rightarrow \frac{9}{a^{2}}-\frac{4 b}{a}=1 \Rightarrow a(a+4 b)=9 \text {. }
$$

Since $a$ and $b$ are both negative integers, we have $a=-1, a+4 b=-9$. Therefore, $a=-1, b=-2$.
(2) Since $\alpha0, \\ 4 a+6+ba+3+b>0$, which means when $x=1$, $a x^{2}+4 x+b>0$.
From $4 a-8+b<4 a+6+b<0$, which means when $x=-2$, $a x^{2}+4 x+b<0$.
Since $a<0$, we have $-2<x_{1}<1<x_{2}$.