19. (12 points) Given vectors
$$
\begin{array}{l}
\boldsymbol{m}=(\sqrt{3} \sin \omega x, \cos \omega x), \\
\boldsymbol{n}=(\cos \omega x,-\cos \omega x),
\end{array}
$$

where $\omega>0$. The function $f(x)=\boldsymbol{m} \cdot \boldsymbol{n}$ has the smallest positive period of $\frac{\pi}{2}$.
(1) Find the value of $\omega$;
(2) In $\triangle ABC$, the sides $a$, $b$, and $c$ satisfy $b^{2}=a c$, and the angle opposite side $b$ is $x$. If the equation $f(x)=k$ has two distinct real solutions for $x$, find the range of real values for $k$.
19. From the given, we have
$$
\begin{array}{l}
f(x)=\frac{\sqrt{3}}{2} \sin 2 \omega x-\cos ^{2} \omega x \\
=\frac{\sqrt{3}}{2} \sin 2 \omega x-\frac{1+\cos 2 \omega x}{2} \\
=\sin \left(2 \omega x-\frac{\pi}{6}\right)-\frac{1}{2} .
\end{array}
$$

Thus, $T=\frac{2 \pi}{2 \omega}=\frac{\pi}{2} \Rightarrow \omega=2$.
(2) Note that,
$$
\begin{array}{l}
\cos x=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{a^{2}+c^{2}-a c}{2 a c} \\
\geqslant \frac{2 a c-a c}{2 a c}=\frac{1}{2} \\
\Rightarrow 0<x \leqslant \frac{\pi}{3} \Rightarrow 4 x-\frac{\pi}{6} \in\left(-\frac{\pi}{6}, \frac{7 \pi}{6}\right) .
\end{array}
$$

Therefore, $f(x)=k \Leftrightarrow \sin \left(4 x-\frac{\pi}{6}\right)=k+\frac{1}{2}$.
From the graph of the function $y=\sin x$, we know that to have two distinct real solutions, we need
$$
-\frac{1}{2}<k+\frac{1}{2}<1 \Rightarrow-1<k<\frac{1}{2} \text {. }
$$