$$
\begin{array}{l}
f(x)=x^{3}+a x^{2}+b x+c, \\
g(x)=3 x^{2}+2 a x+b,
\end{array}
$$

and $\triangle A B C$ satisfies the following conditions:
(1) $\angle A, \angle B, \angle C$ form an arithmetic sequence;
(2) $\tan B$ is a root of the equation $f(x)=0$;
(3) $\tan A, \tan C$ are two distinct real roots of the equation $g(x)=0$;
(4) The function values $f(\tan A)$ and $f(\tan C)$ are opposite numbers.
Find $f(x), g(x)$, and the degree measures of $\angle A, \angle B, \angle C$.
From condition (1), we get $\angle B=\frac{\pi}{3}, \angle A+\angle C=\frac{2 \pi}{3}$.
From condition (2), we get
$$
f(\tan B)=f(\sqrt{3})=3 \sqrt{3}+3 a+\sqrt{3} b+c=0 \text {. }
$$

From condition (3), we get $\Delta=4 a^{2}-12 b=4\left(a^{2}-3 b\right)>0$,
$\tan A+\tan C=-\frac{2 a}{3}, \tan A \cdot \tan C=\frac{b}{3}$.
Thus, $\tan (A+C)=\frac{\tan A+\tan C}{1-\tan A \cdot \tan C}$
$$
=\frac{-\frac{2 a}{3}}{1-\frac{b}{3}}=\tan \frac{2 \pi}{3}=-\sqrt{3} \text {. }
$$

Simplifying, we get $a=-\frac{\sqrt{3}}{2}(b-3)$.
Substituting into equation (1), we get $c=\frac{\sqrt{3}}{2} b-\frac{15 \sqrt{3}}{2}$.
From condition (4), we know $f(\tan A)+f(\tan C)=0$, i.e., $\tan ^{3} A+\tan ^{3} C+a\left(\tan ^{2} A+\tan ^{2} C\right)+$ $b(\tan A+\tan C)+2 c=0$.
Thus, $(\tan A+\tan C)^{3}-3 \tan A \cdot \tan C(\tan A+\tan C)+$ $a\left[(\tan A+\tan C)^{2}-2 \tan A \cdot \tan C\right]+$ $b(\tan A+\tan C)+2 c=0$.
Substituting equation (2) and simplifying, we get
$$
\frac{4}{27} a^{3}-\frac{2}{3} a b+2 c=0 .
$$

Thus, $\frac{4}{27}\left[-\frac{\sqrt{3}}{2}(b-3)\right]^{3}-\frac{2}{3} b\left[-\frac{\sqrt{3}}{2}(b-3)\right]+$
$$
2\left(\frac{\sqrt{3}}{2} b-\frac{15 \sqrt{3}}{2}\right)=0 \text {. }
$$

Simplifying, we get $b^{3}-15 b^{2}+27 b+243=0$, i.e.,
$$
(b+3)(b-9)^{2}=0 \text {. }
$$

Solving, we get $b=-3$ or $b=9$.
Considering condition (3), we have
$$
a=3 \sqrt{3}, b=-3, c=-9 \sqrt{3} \text {. }
$$

Thus, $\tan A \cdot \tan C=\frac{b}{3}=-1$.
Hence, $\tan A=-\cot C=\tan \left(C-\frac{\pi}{2}\right)$.
Therefore, $\angle A=\angle C-\frac{\pi}{2}$ or $\angle A=\angle C-\frac{\pi}{2}+\pi$.
Since $\angle A+\angle C=\frac{2 \pi}{3}$, we have $\angle A=\frac{\pi}{12}, \angle C=\frac{7 \pi}{12}$ or $\angle A=\frac{7 \pi}{12}, \angle C=\frac{\pi}{12}$.

Thus, the three interior angles of the triangle are $\frac{\pi}{12}, \frac{\pi}{3}, \frac{7 \pi}{12}$.
$$
\begin{array}{l}
f(x)=x^{3}+3 \sqrt{3} x^{2}-3 x-9 \sqrt{3}, \\
g(x)=3 x^{2}+6 \sqrt{3} x-3 .
\end{array}
$$