Three. (20 points). Given $a=(5 \cos \alpha, 4 \sin \alpha)$, $\boldsymbol{b}=(5 \cos \beta, 4 \sin \beta)$. Find the maximum value of $|\boldsymbol{a}-\boldsymbol{b}|$.


$$
\begin{array}{l}
\text { III. Solution 1: }|\boldsymbol{a}-\boldsymbol{b}| \\
=\sqrt{5(\cos \alpha-\cos \beta)^{2}+4(\sin \alpha-\sin \beta)^{2}} \\
\leqslant \sqrt{5(\cos \alpha-\cos \beta)^{2}+5(\sin \alpha-\sin \beta)^{2}} \\
=\sqrt{10-10(\cos \alpha \cdot \cos \beta+\sin \alpha \cdot \sin \beta)} \\
=\sqrt{10[1-\cos (\alpha-\beta)]}=\sqrt{20 \sin ^{2} \frac{\alpha-\beta}{2}} \\
=2 \sqrt{5}\left|\sin \frac{\alpha-\beta}{2}\right| \leqslant 2 \sqrt{5} . \\
\text { When } \alpha=\pi, \beta=0 \text {, we have }\left\{\begin{array}{l}
\sin \alpha-\sin \beta=0, \\
\sin \frac{\alpha-\beta}{2}=1,
\end{array}\right.
\end{array}
$$
$|\boldsymbol{a}-\boldsymbol{b}|$ reaches its maximum value $2 \sqrt{5}$.
Solution 2: As shown in Figure 2,

On the ellipse $\frac{x^{2}}{5}+\frac{y^{2}}{4}=1$,
take points
$$
\begin{array}{l}
P(\sqrt{5} \cos \alpha, 2 \sin \alpha) \text {, } \\
Q(\sqrt{5} \cos \beta, 2 \sin \beta) . \\
\text { Using the major axis of the ellipse }
\end{array}
$$

Using the major axis of the ellipse
$A_{1} A_{2}=2 \sqrt{5}$ as the diameter, draw a circle $x^{2}+y^{2}=5$, intersecting the line $P Q$ at points $P_{1}$ and $Q_{1}$.
For any point $(x, y)$ on the circle (including $P_{1}$ and $Q_{1}$), we have
$$
\frac{x^{2}}{5}+\frac{y^{2}}{4} \geqslant \frac{x^{2}}{5}+\frac{y^{2}}{5}=1 \text {, }
$$

which means $P_{1}$ and $Q_{1}$ are outside the ellipse, so $P Q \leqslant P_{1} Q_{1}$.
Since the diameter is the largest chord in a circle, $P_{1} Q_{1} \leqslant A_{1} A_{2}$. Therefore, $P Q \leqslant P_{1} Q_{1} \leqslant A_{1} A_{2}=2 \sqrt{5}$, i.e.,
$$
\begin{array}{l}
|\boldsymbol{a}-\boldsymbol{b}|=\sqrt{5(\cos \alpha-\cos \beta)^{2}+4(\sin \alpha-\sin \beta)^{2}} \\
\quad \leqslant 2 \sqrt{5} .
\end{array}
$$

When $P Q$ is $A_{1} A_{2}$, $|\boldsymbol{a}-\boldsymbol{b}|$ reaches its maximum value $2 \sqrt{5}$.
Solution 3: More generally, as shown in Figure 3, let $F_{1}$ and $F_{2}$ be the foci of the ellipse. By the triangle inequality,
$$
\begin{aligned}
P Q= & \frac{1}{2}(P Q+P Q) \\
\leqslant & \frac{1}{2}\left[\left(\left|P F_{1}\right|+\left|Q F_{1}\right|\right)+\text { Figure } 3\right. \\
& \left.\left(\left|P F_{2}\right|+\left|Q F_{2}\right|\right)\right] \\
= & \frac{1}{2}\left[\left(\left|P F_{1}\right|+\left|P F_{2}\right|\right)+\left(\left|Q F_{1}\right|+\left|Q F_{2}\right|\right)\right] \\
= & \frac{1}{2}(2 a+2 a)=2 a .
\end{aligned}
$$

When $P, Q, F_{1}, F_{2}$ are collinear, $P Q$ is $A_{1} A_{2}$, reaching its maximum value $2 a$.

Note: This problem is based on the fact that "the major axis is the largest chord of the ellipse," and there can be multiple solutions using algebra, geometry, and trigonometry.
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