Sure, here is the translated text:

```
II. Fill-in-the-Blanks (Total 54 points, 9 points per question)
1. If $x^{2}-x+2$ is a factor of $a x^{9}+b x^{8}+1$, then $a=$ $\qquad$ .
```
Ni. $1 . a=-\frac{3}{256}$.
Let the roots of $x^{2}-x+2=0$ be $x_{1} 、 x_{2}\left(x_{1}+x_{2}=1\right.$. $x_{1} x_{2}=2$ )
From $\left\{\begin{array}{l}a x_{1}^{9}+b x_{1}^{8}+1=0, \\ a x_{2}^{9}+b x_{2}^{8}+1=0\end{array}\right.$

Eliminating $b$ gives
$$
\begin{array}{l}
a x_{1}^{8} x_{2}^{8}\left(x_{1}-x_{2}\right)=x_{1}^{8}-x_{2}^{8} . \\
\text { Hence } a \cdot 2^{8}=\left(x_{1}^{4}+x_{2}^{4}\right)\left(x_{1}^{2}+x_{2}^{2}\right)\left(x_{1}+x_{2}\right) \\
\quad=\left(x_{1}^{4}+x_{2}^{4}\right)\left(x_{1}^{2}+x_{2}^{2}\right) \cdot 1 . \\
\because x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=-3, \\
x_{1}^{4}+x_{2}^{4}=\left(x_{1}^{2}+x_{2}^{2}\right)^{2}-2 x_{1}^{2} x_{2}^{2}=1, \\
\therefore a=\frac{-3}{2^{8}}=-\frac{3}{256} .
\end{array}
$$