16. Given the even function
$$
\begin{aligned}
f(x)= & 5 \cos \theta \cdot \sin x-5 \sin (x-\theta)+ \\
& (4 \tan \theta-3) \sin x-5 \sin \theta
\end{aligned}
$$

has a minimum value of -6.
(1) Find the maximum value of $f(x)$ and the set of $x$ at which this maximum occurs.
(2) Let the function
$$
g(x)=\lambda f(\omega x)-f\left(\omega x+\frac{\pi}{2}\right)(\lambda>0, \omega>0) .
$$

It is known that $y=g(x)$ attains its minimum value at $x=\frac{\pi}{6}$ and the point $\left(\frac{2 \pi}{3}, 3-3 \lambda\right)$ is a center of symmetry of its graph. Try to find the minimum value of $\lambda+\omega$.
16. (1) Simplify to get
$$
f(x)=5 \cos x \cdot \sin \theta+(4 \tan \theta-3) \sin x-5 \sin \theta \text {. }
$$

Since $f(x)$ is an even function, we have
$(4 \tan \theta-3) \sin x=0$
for all $x \in \mathbf{R}$.
Thus, $4 \tan \theta-3=0, \tan \theta=\frac{3}{4}$.
Therefore, $f(x)=5 \cos x \cdot \sin \theta-5 \sin \theta$
$$
=5 \sin \theta(\cos x-1) \text {, }
$$

with its minimum value being -6. At this point, we have
$$
\sin \theta=\frac{3}{5}, \cos x=-1 \text {. }
$$

Thus, $f(x)=3(\cos x-1)$.
Therefore, the maximum value of $f(x)$ is 0, at which point, the set of $x$ is $\{x \mid x=2 k \pi, k \in \mathbf{Z}\}$.
$$
\begin{array}{l}
\text { (2) } g(x)=\lambda f(\omega x)-f\left(\omega x+\frac{\pi}{2}\right) \\
=3 \lambda \cos \omega x-3 \lambda-3 \cos \left(\omega x+\frac{\pi}{2}\right)+3 \\
=3 \lambda \cos \omega x-3 \lambda+3 \sin \omega x+3 .
\end{array}
$$

Since $g(x)$ has a minimum value at $x=\frac{\pi}{6}$, the graph of $g(x)$ is symmetric about $x=\frac{\pi}{6}$, so
$$
g\left(-\frac{\pi}{3}\right)=g\left(\frac{2 \pi}{3}\right)=3-3 \lambda .
$$

Thus, $3 \lambda \cos \left(-\frac{\omega \pi}{3}\right)+3 \sin \left(-\frac{\omega \pi}{3}\right)=0$,
and $3 \lambda \cos \frac{2 \omega \pi}{3}+3 \sin \frac{2 \omega \pi}{3}=0$.
Therefore, $\lambda=\tan \frac{\omega \pi}{3}=-\tan \frac{2 \omega \pi}{3}$
$=\tan \left(-\frac{2 \omega \pi}{3}\right)=\tan \left(k \pi-\frac{2 \omega \pi}{3}\right)$.
Then $\frac{\omega \pi}{3}=k \pi-\frac{2 \omega \pi}{3}$, i.e., $\omega=k \in \mathbf{Z}$.
Since $\omega>0$, $\omega$ is a positive integer.
Because $\lambda>0$ and $\omega$ is a positive integer, we have $\omega=3 l-2$ (where $l$ is a positive integer), $\lambda=\sqrt{3}$.
When $\omega=1$,
$$
g(x)=3 \sqrt{3} \cos x+3 \sin x+3-3 \sqrt{3} \text {. }
$$

Clearly, $g(x)$ has a maximum value at $x=\frac{\pi}{6}$, not a minimum value. Contradiction.
When $\omega=4$,
$$
g(x)=3 \sqrt{3} \cos 4 x+3 \sin 4 x+3-3 \sqrt{3} \text {. }
$$

Clearly, $g(x)$ has neither a maximum nor a minimum value at $x=\frac{\pi}{6}$. Contradiction.
When $\omega=7$,
$$
g(x)=3 \sqrt{3} \cos 7 x+3 \sin 7 x+3-3 \sqrt{3} \text {. }
$$

Clearly, $g(x)$ has a minimum value at $x=\frac{\pi}{6}$, and the graph of $y=g(x)$ is symmetric about the point $\left(\frac{2 \pi}{3}, 3-3 \sqrt{3}\right)$.
Therefore, the minimum value of $\lambda+\omega$ is $\sqrt{3}+7$.