8. (2004 National College Entrance Examination, Liaoning Province) Given the function $f(x)=\ln \left(\mathrm{e}^{x}+a\right)(a>0)$.
(I) Find the inverse function $y=f^{-1}(x)$ of the function $y=f(x)$ and the derivative $f^{\prime}(x)$ of $f(x)$;
(II) Assuming that for any $x \in[\ln (3 a), \ln (4 a)]$, the inequality $\left|m-f^{-1}(x)\right|+\ln \left[f^{\prime}(x)\right]<0$ holds, find the range of the real number $m$.

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8. ( I ) From $y=f(x)=\ln \left(\mathrm{e}^{x}+a\right)$, we get $x=\ln \left(\mathrm{e}^{y}-a\right)$.

Then $y=f^{-1}(x)=\ln \left(\mathrm{e}^{x}-a\right)(x>\ln a)$.
$$
f^{\prime}(x)=\frac{\mathrm{e}^{x}}{\mathrm{e}^{x}+a} \text {. }
$$
(II) Solution: From $\left|m-f^{-1}(x)\right|+\left|\ln \left(f^{\prime}(x)\right)\right|0, \\
v\left(t_{2}\right)-v\left(t_{1}\right)=\cdots=\frac{t_{1} t_{2}\left(t_{2}-t_{1}\right)+a^{2}\left(t_{2}-t_{1}\right)}{t_{1} t_{2}}>0,
\end{array}
$$

Thus, $u(t)$ and $v(t)$ are increasing functions on $[3 a, 4 a]$.
Hence $u(t)_{\operatorname{max}}=u(4 a)=\frac{12}{5} a, v(t)_{\min }=v(3 a)=\frac{8}{3} a$.
Therefore, $\frac{12}{5} a0, \psi^{\prime}(x)>0$, thus $\varphi(x)$ and $\psi(x)$ are increasing functions on $[\ln (3 a), \ln (4 a)]$.
Hence $\varphi(\ln (4 a))<m<\psi(\ln (3 a))$,
which means $\ln \left(\frac{12}{5} a\right)<m<\ln \left(\frac{8}{3} a\right)$.