9. (20 points) Given $O$ as the origin, $N(1,0)$, and $M$ as a moving point on the line $x=-1$, the angle bisector of $\angle M O N$ intersects the line $M N$ at point $P$.
(1) Find the equation of the trajectory curve $E$ of point $P$;
(2) Draw a line $l$ with slope $k$ through the point $Q\left(-\frac{1}{2},-\frac{1}{2}\right)$. If the line $l$ intersects the curve $E$ at exactly one point, find the range of values for $k$.
(1) Let $P(x, y), M(-1, t)$.
It is easy to know, $0 \leqslant x<1$.
Since $O P$ bisects $\angle M O N$, we have
$\overrightarrow{M P}=\frac{|O M|}{|O N|} \overrightarrow{P N}=\sqrt{1+t^{2}} \overrightarrow{P N}$.
Then $x+1=\sqrt{1+t^{2}}(1-x)$,
$y-t=\sqrt{1+t^{2}}(0-y)$.
From equations (1) and (2), we get $t=\frac{2 y}{1-x}$.
Substituting into equation (1) yields $\frac{x+1}{1-x}=\sqrt{1+\left(\frac{2 y}{1-x}\right)^{2}}$.
Simplifying, we obtain the curve $E: y^{2}=x(0 \leqslant x<1)$.
(2) Let $A(1,1), B(1,-1)$. Then
$k_{Q A}=1, k_{Q B}=-\frac{1}{3}$.
Thus, the line $l: y+\frac{1}{2}=k\left(x+\frac{1}{2}\right)$.
Combining with the equation of curve $E$ gives
$k y^{2}-y+\frac{1}{2}(k-1)=0$.
When the line $l$ is tangent to the parabola $y^{2}=x$ at point $T$,
$$
\Delta=1-2 k(k-1)=0 \Rightarrow k_{1,2}=\frac{1 \pm \sqrt{3}}{2} \text {. }
$$

When $k=k_{1}=\frac{1+\sqrt{3}}{2}$, $y_{T}=\frac{\sqrt{3}-1}{2}$, the point $T$ is on the curve $E$;
When $k=k_{2}=\frac{1-\sqrt{3}}{2}$, $y_{T}=-\frac{\sqrt{3}+1}{2}$, the point $T$ is not on the curve $E$.
If the line $l$ intersects the curve $E$ at exactly one point, then $k_{Q B}<k \leqslant k_{Q A}$ or $k=\frac{1+\sqrt{3}}{2}$.
Therefore, $k \in\left(-\frac{1}{3}, 1\right] \cup\left\{\frac{1+\sqrt{3}}{2}\right\}$.