8. Given that the angle between skew lines $a$ and $b$ is $60^{\circ}$, through a point $P$ in space, draw a line $l$ that makes an angle $\alpha$ $\left(0^{\circ}<\alpha<90^{\circ}\right)$ with both $a$ and $b$, then the number of such lines $l$ is $f(\alpha)=$ $\qquad$
8. $f(\alpha)=\left\{\begin{array}{l}0\left(0^{\circ}<\alpha<30^{\circ}\right) \\ 1\left(\alpha=30^{\circ}\right) \\ 2\left(30^{\circ}<\alpha<60^{\circ}\right) \\ 3\left(\alpha=60^{\circ}\right) \\ 4\left(60^{\circ}<\alpha<90^{\circ}\right)\end{array}\right.$

Draw lines $a', b'$ through point $P$ parallel to lines $a, b$, respectively. Then lines $a', b'$ intersect at angles of $60^{\circ}$ and $120^{\circ}$. Therefore, when $0^{\circ}<\alpha<30^{\circ}$, line $l$ does not exist; when $\alpha=30^{\circ}$, line $l$ can only be the angle bisector of the angle between lines $a'$ and $b'$, so there is 1 line; when $30^{\circ}<\alpha<60^{\circ}$, through point $P$ in space, two lines $l$ can be drawn that form an angle $\alpha$ with both lines $a'$ and $b'$; when $\alpha=60^{\circ}$, line $l$ can be the angle bisector of the $120^{\circ}$ angle, plus the two lines from the previous case, making a total of 3 lines; when $60^{\circ}<\alpha<90^{\circ}$, through point $P$ in space, 4 lines $l$ can be drawn that form an angle $\alpha$ with both lines $a'$ and $b'$.
9. $\left[-\frac{1}{2}, 0\right) \cup\left(0, \frac{45}{8}\right)$
Clearly, $x \geqslant-\frac{1}{2}, x \neq 0$, the original inequality simplifies to: $x^{2}(8 x-45)<0$, thus $x<\frac{45}{8}$,
so $-\frac{1}{2} \leqslant x<0$ or $0<x<\frac{45}{8}$.