II. (50 points) Find the maximum value of the area of an inscribed triangle in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$.
As shown in Figure 7, let \( A\left(a \cos \theta_{1}, b \sin \theta_{1}\right) \), \( B\left(a \cos \theta_{2}, b \sin \theta_{2}\right) \), \( C\left(a \cos \theta_{3}, b \sin \theta_{3}\right) \), where \( 0 \leqslant \theta_{1} < \theta_{2} < \theta_{3} < 2 \pi \). Other cases are similar.
Draw the line \( l: y = -b \).
The projections of points \( A \), \( B \), and \( C \) on \( l \) are \( A_{1} \), \( B_{1} \), and \( C_{1} \) respectively, with coordinates
\[
\begin{array}{l}
=\frac{1}{2}\left(\left|A A_{1}\right|+\left|B B_{1}\right|\right) \cdot\left|A_{1} B_{1}\right|+\frac{1}{2}\left(\left|B B_{1}\right|+\right. \\
\left.\left|C C_{1}\right|\right) \cdot\left|B_{1} C_{1}\right|-\frac{1}{2}\left(\left|A A_{1}\right|+\left|C C_{1}\right|\right) \cdot\left|A_{1} C_{1}\right| \\
=\frac{1}{2}\left(b \sin \theta_{1}+b+b \sin \theta_{2}+b\right)\left(a \cos \theta_{1}-a \cos \theta_{2}\right)+ \\
\frac{1}{2}\left(b \sin \theta_{2}+b+b \sin \theta_{3}+b\right)\left(a \cos \theta_{2}-a \cos \theta_{3}\right)- \\
\frac{1}{2}\left(b \sin \theta_{1}+b+b \sin \theta_{3}+b\right)\left(a \cos \theta_{1}-a \cos \theta_{3}\right) \\
=\frac{1}{2} a b\left[\sin \left(\theta_{2}-\theta_{1}\right)+\sin \left(\theta_{3}-\theta_{2}\right)-\sin \left(\theta_{3}-\theta_{1}\right)\right] \text{. } \\
\end{array}
\]

Since \( 0 \leqslant \theta_{1} < \theta_{2} < \theta_{3} < 2 \pi \), then
\[
\theta_{2}-\theta_{1}, \theta_{3}-\theta_{2}, \theta_{3}-\theta_{1} \in (0,2 \pi) \text{. }
\]

Let \( \alpha = \theta_{2} - \theta_{1} \), \( \beta = \theta_{3} - \theta_{2} \), then
\[
\alpha + \beta = \theta_{3} - \theta_{1} \in (0,2 \pi) \text{. }
\]

Thus, the problem is transformed into:
When \( \alpha, \beta, \alpha + \beta \in (0,2 \pi) \), find the maximum value of \( \sin \alpha + \sin \beta - \sin (\alpha + \beta) \).
\[
\begin{array}{l}
P = \sin \alpha + \sin \beta - \sin (\alpha + \beta) \\
= 2 \sin \frac{\alpha + \beta}{2} \cdot \cos \frac{\alpha - \beta}{2} - 2 \sin \frac{\alpha + \beta}{2} \cdot \cos \frac{\alpha + \beta}{2} \\
= 2 \sin \frac{\alpha + \beta}{2} \left( \cos \frac{\alpha - \beta}{2} - \cos \frac{\alpha + \beta}{2} \right) \\
\leqslant 2 \sin \frac{\alpha + \beta}{2} \left( 1 - \cos \frac{\alpha + \beta}{2} \right) \\
= 4 \sin \frac{\alpha + \beta}{2} \cdot \sin^2 \frac{\alpha + \beta}{4} = 8 \sin^3 \frac{\alpha + \beta}{4} \cdot \cos \frac{\alpha + \beta}{4} . \\
P^2 \leqslant 64 \sin^6 \frac{\alpha + \beta}{4} \cos^2 \frac{\alpha + \beta}{4} \\
= \frac{64}{3} \sin^2 \frac{\alpha + \beta}{4} \cdot \sin^2 \frac{\alpha + \beta}{4} \cdot \sin^2 \frac{\alpha + \beta}{4} \cdot 3 \cos^2 \frac{\alpha + \beta}{4} \\
\leqslant \frac{64}{3} \left( \frac{\sin^2 \frac{\alpha + \beta}{4} + \sin^2 \frac{\alpha + \beta}{4} + \sin^2 \frac{\alpha + \beta}{4} + 3 \cos^2 \frac{\alpha + \beta}{4}}{4} \right)^4 = \frac{\beta^3}{4} .
\end{array}
\]

Therefore, \( P \leqslant \frac{3 \sqrt{3}}{2} \).
The equality holds if and only if \( \left\{ \begin{array}{l} \cos \frac{\alpha - \beta}{2} = 1, \\ \sin^2 \frac{\alpha + \beta}{4} = 3 \cos^2 \frac{\alpha + \beta}{4}, \end{array} \right. \) i.e., \( \alpha = \beta = \frac{2 \pi}{3} \).
Thus, \( \left(S_{\triangle A B C}\right)_{\max} = \frac{3 \sqrt{3}}{4} a b \).