I3.1 Let $x \neq \pm 1$ and $x \neq-3$. If $a$ is the real root of the equation $\frac{1}{x-1}+\frac{1}{x+3}=\frac{2}{x^{2}-1}$, find the value of $a$
I3.2 If $b>1, f(b)=\frac{-a}{\log _{2} b}$ and $g(b)=1+\frac{1}{\log _{3} b}$. If $b$ satisfies the equation $|f(b)-g(b)|+f(b)+g(b)=3$, find the value of $b$.

I3.3 Given that $x_{0}$ satisfies the equation $x^{2}-5 x+(b-8)=0$. If $c=\frac{x_{0}^{2}}{x_{0}^{4}+x_{0}^{2}+1}$, find the value of $c$.
$$
\begin{array}{l}
\frac{1}{x+3}=\frac{2}{x^{2}-1}-\frac{1}{x-1} \Rightarrow \frac{1}{x+3}=\frac{2-(x+1)}{(x-1)(x+1)} \\
\frac{1}{x+3}=\frac{1-x}{(x-1)(x+1)} \Rightarrow \frac{1}{x+3}=-\frac{1}{x+1} \\
x+1=-x-3 \Rightarrow x=-2=a \\
\left|\frac{2 \log 2}{\log b}-1-\frac{\log 3}{\log b}\right|+\frac{2 \log 2}{\log b}+1+\frac{\log 3}{\log b}=3 \Rightarrow\left|\frac{\log 4 / 3 b}{\log b}\right|+\frac{\log 12}{\log b}=2 \Rightarrow\left|\log \frac{4}{3 b}\right|=\log \frac{b^{2}}{12} \\
\log \frac{4}{3 b}= \pm \log \frac{b^{2}}{12} \\
\log \frac{4}{3 b}=\log \frac{b^{2}}{12} \text { or } \log \frac{4}{3 b}=\log \frac{12}{b^{2}} \\
b^{3}=16 \text { or } b=9
\end{array}
$$

When $b^{3}=16,\left(b^{3}\right)^{2}=2560$ accepted.
Method 2
Remark Define the maximum function of $x, y$ as: $\operatorname{Max}(x, y)=\frac{x+y+|x-y|}{2}$
Similarly, the minimum function of $x, y$ is: $\operatorname{Min}(x, y)=\frac{x+y-|x-y|}{2}$
$$
f(b)=\frac{-a}{\log _{2} b}=\frac{2 \log 2}{\log b}=\frac{\log 4}{\log b}, g(b)=1+\frac{1}{\log _{3} b}=\frac{\log b}{\log b}+\frac{\log 3}{\log b}=\frac{\log 3 b}{\log b}
$$

The given equation is equivalent to $2 \operatorname{Max}(f(b), g(b))=3$
If $f(b)>g(b)$, i.e. $\frac{\log 4}{\log b}>\frac{\log 3 b}{\log b} \Rightarrow b<\frac{4}{3}$, then the equation is $2 f(b)=3$
$$
\frac{2 \log 4}{\log b}=3 \Rightarrow \log 16=\log b^{3} \Rightarrow b^{3}=16<\left(\frac{4}{3}\right)^{3} \Rightarrow 1<\frac{4}{27}
$$
$\Rightarrow 27<4$, which is a contradiction; $\therefore$ rejected
If $f(b) \leq g(b)$, the equation is equivalent to $2 g(b)=3$
$$
\begin{array}{l}
\text { i.e. } \frac{2 \log 3 b}{\log b}=3 \Rightarrow \log 9 b^{2}=\log b^{3} \Rightarrow 9 b^{2}=b^{3} \Rightarrow b=9 \\
x^{2}-5 x+1=0, x^{2}+1=5 x, x^{4}+2 x^{2}+1=25 x^{2}, x^{4}+x^{2}+1=24 x^{2} \\
c=\frac{x_{0}^{2}}{x_{0}^{4}+x_{0}^{2}+1}=\frac{x_{0}^{2}}{24 x_{0}^{2}}=\frac{1}{24}
\end{array}
$$