Five. (20 points) As shown in Figure 2, in the Cartesian coordinate system, a line segment $P Q$ of length 6 has one endpoint $P$ sliding on the ray $y=0(x \leqslant 0)$, and the other endpoint $Q$ sliding on the ray $x=0(y \leqslant 0)$. Point $M$ is on line segment $P Q$, and $\frac{P M}{M Q}=\frac{1}{2}$.
(1) Find the equation of the trajectory of point $M$;
(2) If the trajectory of point $M$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, find the maximum area of quadrilateral $O A M B$ (where $O$ is the origin).
(1) Let $P\left(x_{1}, 0\right)$, $Q\left(0, y_{1}\right)$, and $M(x, y)$, where $x \leqslant 0$ and $y_{1} \leqslant 0$.
From the conditions, we have
$$
x_{1}^{2}+y_{1}^{2}=36 \text{. }
$$

Also, given $\lambda=\frac{P M}{M Q}=\frac{1}{2}$, we get
$$
x=\frac{x_{1}}{1+\frac{1}{2}}, y=\frac{\frac{1}{2} y_{1}}{1+\frac{1}{2}} \text{. }
$$

Substituting $x_{1}=\frac{3}{2} x$ and $y_{1}=3 y$ into equation (1), we obtain
$$
\frac{x^{2}}{16}+\frac{y^{2}}{4}=1(x, y \leqslant 0) \text{. }
$$

Therefore, the equation of the trajectory of point $M$ is
$$
\frac{x^{2}}{16}+\frac{y^{2}}{4}=1(x \leqslant 0, y \leqslant 0) \text{. }
$$
(2) As shown in Figure 5, let point
$M(4 \cos \alpha, 2 \sin \alpha)$, where $0 \leqslant \alpha < 2 \pi$.
From the conditions $\left\{\begin{array}{l}4 \cos \alpha<0, \\ 2 \sin \alpha<0\end{array}\right.$, we get
$$
\pi<\alpha<\frac{3 \pi}{2} \text{. }
$$

Thus, $S_{\text {quadrilateral, MMB }}=S_{\triangle M M}+S_{\triangle O M M}$.
And $S_{\triangle M M}=\frac{1}{2}|O A| \cdot\left|y_{M}\right|=\frac{1}{2} \times 4 \times|2 \sin \alpha|$ $=-4 \sin \alpha$,
$S_{\text {ΔOEM }}=\frac{1}{2}|O B| \cdot\left|x_{M}\right|=\frac{1}{2} \times 2 \times|4 \cos \alpha|$
$$
=-4 \cos \alpha \text{, }
$$

Therefore, $S_{\text {quadrilateral } 04 m B}=-4(\sin \alpha+\cos \alpha)$
$$
=-4 \sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right) \leqslant 4 \sqrt{2} \text{. }
$$

The maximum area of quadrilateral $O A M B$ is $4 \sqrt{2}$ when and only when $\alpha=\frac{5 \pi}{4} \in\left(\pi, \frac{3 \pi}{2}\right)$.