II. (50 points)
Make a trigonometric substitution. Since $|a|>1$, we can set
$$
a=\frac{1}{\cos \theta}(\theta \in(0,2 \pi)) \text {. }
$$

Then $b=\frac{a^{2}}{2-a^{2}}=\frac{\frac{1}{\cos ^{2} \theta}}{2-\frac{1}{\cos ^{2} \theta}}=\frac{1}{\cos ^{2} \theta}$,
Similarly, $c=\frac{1}{\cos ^{4} \theta}, a=\frac{1}{\cos ^{8} \theta}$.
Thus, $\cos 8 \theta=\cos \theta, \cos \theta-\cos 8 \theta=0$. And $\cos \theta \neq 0, \cos \theta \neq \pm 1$,
$$
\begin{array}{l}
2 \sin \frac{7}{2} \theta \cdot \sin \frac{9}{2} \theta=0 . \\
\frac{7}{2} \theta=k \pi \quad k \in \mathbb{Z} . \\
\frac{9}{2} \theta=k \pi \quad k \in \mathbb{Z} .
\end{array}
$$

Since we only need to consider the value of $\cos \theta$,
$\cos \theta$ can only be $\cos \frac{4}{7} \pi, -\cos \frac{\pi}{7}, \cos \frac{2}{7} \pi, \cos \frac{4}{9} \pi, -\cos \frac{\pi}{9}, -\cos \frac{2}{9} \pi$.
1. When $\cos \theta=\cos \frac{4}{7} \pi$ or $-\cos \frac{\pi}{7}$ or $\cos \frac{2}{7} \pi$:
$$
\begin{array}{l}
a+b+c=\frac{1}{\cos \frac{4}{7} \pi}+\frac{1}{\cos \frac{8}{7} \pi}+\frac{1}{\cos \frac{16}{7} \pi} \\
=\frac{1}{\cos \frac{4}{7} \pi}+\frac{1}{-\cos \frac{\pi}{7}}+\frac{1}{\cos \frac{2}{7} \pi} \\
=\frac{2}{\varepsilon^{2}+\frac{1}{\varepsilon^{2}}}+\frac{2}{\varepsilon^{4}+\frac{1}{\varepsilon^{4}}}+\frac{2}{\varepsilon^{8}+\frac{1}{\varepsilon^{8}}} \\
\left(\text { Let } \varepsilon=\cos \frac{2}{7} \pi+i \sin \frac{2}{7} \pi\right) \\
=2 \cdot\left(\frac{\varepsilon^{2}}{\varepsilon^{4}+1}+\frac{\varepsilon^{4}}{\varepsilon+1}+\frac{\varepsilon}{\varepsilon^{2}+1}\right) \\
=2 \cdot \frac{-2 \varepsilon^{7}}{\varepsilon^{7}}=-4 .
\end{array}
$$
2. When $\cos \theta=\cos \frac{4}{9} \pi$ or $-\cos \frac{\pi}{9}$ or $\cos \frac{2}{9} \pi$:
$$
\begin{array}{l}
a+b+c=\frac{1}{\cos \frac{4}{9} \pi}+\frac{1}{\cos \frac{8}{9} \pi}+\frac{1}{\cos \frac{16}{9} \pi} \\
=\frac{2}{z^{2}+\frac{1}{z^{2}}}+\frac{2}{z^{4}+\frac{1}{z^{4}}}+\frac{2}{z^{8}+\frac{1}{z^{8}}}
\end{array}
$$
(Let $\left.z=\cos \frac{2}{9} \pi+i \sin \frac{2}{9} \pi\right)$
$$
\begin{array}{l}
=2\left(\frac{z^{2}}{z^{4}+1}+\frac{z^{4}}{z^{8}+1}+\frac{z^{8}}{z^{16}+1}\right) \\
=2 \cdot \frac{2 z^{2}+2 z^{8}-z^{5}}{-z^{5}} \\
=2 \cdot\left(\frac{2}{-z^{3}}+\frac{2}{-z^{-3}}+1\right)=6 .
\end{array}
$$
3. When $\cos \theta=-\cos \frac{\pi}{3}$:
$$
a+b+c=\frac{1}{\cos \frac{4}{3} \pi}+\frac{1}{\cos \frac{8}{3} \pi}+\frac{1}{\cos \frac{12}{3} \pi}=-6 .
$$

Therefore, the possible values of $a+b+c$ are $6, -4, -6$.
II. (50 points $\}$  
$a, b, c \in \mathbf{R}$. Satisfy $|a|>1,|b|>1,|c|>1$ and $b=\frac{a^{2}}{2-a^{2}}, c=\frac{b^{2}}{2-b^{2}}, a=$ $\frac{c^{2}}{2-c^{2}}$. Find all possible values of $a+b+c$.