Nine, for the inequality $\frac{1}{2} x^{2}-3 x+m+8>0$ about $\mathrm{x}$, the solution set is all real numbers. For the equation $(m+3) x^{2}+m x+\frac{2 m+3}{8}=0$ about $x$, the two real roots are $\alpha$ and $\beta$. When $m$ is an integer and $\alpha \neq \beta$, find the value of $\alpha \beta^{4}+\alpha^{4}$.

(10 points)
Nine, solve $\because a=\frac{1}{2}, \frac{1}{2} x^{2}-3 x+m+8>0$ of
the solution set is all real numbers, $\therefore$ the parabola $y=\frac{1}{2} x^{2}-3 x+m+8$ opens
upward, does not intersect the x-axis, i.e., $\triangle-\frac{7}{2} \\
\because \alpha \neq r^{3}, \text { i.e., the equation } (m+3) x^{2}+mx \\
\quad+\left(\frac{2 m+3}{8}\right)=0 \text { has two distinct real roots, } \\
\therefore m+3 \neq 0, \text { and } \triangle' = m^{2}-4(m+3)\left(\frac{2 m+3}{8}\right) \\
\quad>0 .
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$$

Solving this inequality gives $m<-1$.
The range of $m$ that satisfies the above conditions is $-\frac{7}{2}<m<-1$. And $m \neq-3$, the integer value in this range is $m=-2$.
$\therefore$ The original equation is $x^{2}-2 x-\frac{1}{8}=0$.
By Vieta's formulas, $\alpha+\beta=2, \alpha \beta=-\frac{1}{8}$,
$$
\begin{array}{l}
\therefore \alpha \beta^{4}+\alpha^{4} \beta=\alpha \beta\left(\alpha^{3}+\beta^{3}\right) \\
=\alpha \beta\left[(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)\right] \\
=-\frac{1}{8}\left[2^{3}-3\left(-\frac{1}{8}\right) \cdot 2\right] \\
=-\frac{1}{8}\left(8+\frac{3}{4}\right)=-1-\frac{3}{32} \\
=-1 \frac{3}{32} .
\end{array}
$$
(By Wu Yizhi)