已知函数 $f(x)=\ln \frac{x+8}{2-x}$ 的定义城为 $D$, 对于任意 $x_{1}, x_{2} \in D$, 当 $\left|x_{1}-x_{2}\right|=2$ 时, $\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|$ 的最小值为
因为 $(2-x)(x+8)>0$, 所以 $f(x)$ 的定义域为 $(-8,2)$,

不妨设 $x_{1}<x_{2}$, 因为 $\left|x_{1}-x_{2}\right|=2$, 所以 $x_{2}=x_{1}+2, x_{1} \in(-8,0)$,

$$
\begin{aligned}
& \left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|=\left|f\left(x_{1}+2\right)-f\left(x_{1}\right)\right|=\ln \frac{x_{1}+2+8}{2-x_{1}-2}-\ln \frac{x_{1}+8}{2-x_{1}}|=| \ln \left(\frac{x_{1}+10}{-x_{1}} \cdot \frac{2-x_{1}}{x_{1}+8}\right) \\
& =\left|\ln \left(\frac{x_{1}^{2}+8 x_{1}-20}{x_{1}^{2}+8 x_{1}}\right)\right|=\left\lvert\, \ln \left(1-\frac{20}{x_{1}^{2}+8 x_{1}}\right)\right.,
\end{aligned}
$$

故当 $x_{1}^{2}+8 x_{1}$ 取得最小值时, 即 $x_{1}=-4$ 时, 此时 $\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|_{\text {取得最小值, }}$

即 $\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|_{\min }=\left|\ln \left(1-\frac{20}{(-4)^{2}+8 \cdot(-4)}\right)\right|=\ln \frac{9}{4}=2 \ln \frac{3}{2}$.

故答案为: $2 \ln \frac{3}{2}$

【分析】不妨设 $x_{1}<x_{2}$, 得 $x_{2}=x_{1}+2, x_{1} \in(-8,0)$, 化简 $\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|$, 根据对数函数单调性即可求解.

三、解答题