1. Let $x_{1}, x_{2}, x_{3}$ be the roots of the equation
$$
x^{3}-17 x-18=0
$$

with $-4<x_{1}<-3$, and $4<x_{3}<5$.
(1) Find the integer part of $x_{2}$;
(2) Find the value of $\arctan x_{1}+\arctan x_{2}+\arctan x_{3}$.
Three, 1. By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}=0, \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=-17, \\
x_{1} x_{2} x_{3}=18 .
\end{array}
$$
(1) Given $x_{1} \in(-4,-3), x_{3} \in(4,5)$, we know $x_{2}=-x_{1}-x_{3} \in(-2,0)$.
Notice that, $f(x)=x^{3}-17 x-18$ satisfies
$$
f(-1)=-20 \text {. }
$$

Thus, $f(x)$ has a root in the interval $(-2,-1)$, i.e., $x_{2} \in(-2,-1)$.
Therefore, the integer part of $x_{2}$ is -2.
(2) Let $\arctan x_{i}=\theta_{i}(i=1,2,3)$.

From (1), we know
$$
\theta_{1} 、 \theta_{2} \in\left(-\frac{\pi}{2},-\frac{\pi}{4}\right), \theta_{3} \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \text {. }
$$

Thus, $\theta_{1}+\theta_{2}+\theta_{3} \in\left(-\frac{3 \pi}{4}, 0\right)$.
Notice that,
$$
\begin{array}{l}
\tan \left(\theta_{1}+\theta_{2}\right) \\
=\frac{\tan \theta_{1}+\tan \theta_{2}}{1-\tan \theta_{1} \cdot \tan \theta_{2}}=\frac{x_{1}+x_{2}}{1-x_{1} x_{2}} .
\end{array}
$$

Therefore, $\tan \left(\theta_{1}+\theta_{2}+\theta_{3}\right)$
$$
\begin{array}{l}
=\frac{\tan \left(\theta_{1}+\theta_{2}\right)+\tan \theta_{3}}{1-\tan \left(\theta_{1}+\theta_{2}\right) \cdot \tan \theta_{3}} \\
=\frac{x_{1}+x_{2}+x_{3}-x_{1} x_{2} x_{3}}{1-\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\right)}=-1 .
\end{array}
$$

This indicates that, $\theta_{1}+\theta_{2}+\theta_{3}=-\frac{\pi}{4}$, i.e., the answer is $-\frac{\pi}{4}$.