Three. (15 points) As shown in Figure 2, a moving point $P$ is on the arc of a segment with $A B=1$ as the chord, and the segment contains a bow angle of $\frac{2 \pi}{3}$ (including the endpoints). Let $A P=x$, $B P=y$, and try to determine the maximum and minimum values of $k=3 x+2 y$.

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In $\triangle A P B$, by the cosine rule we have
$$
x^{2}+y^{2}-2 x y \cos \frac{2 \pi}{3}=1 \text {, }
$$

which simplifies to $x^{2}+y^{2}+x y=1$.
Given $k=3 x+2 y$, we have
$$
y=\frac{k-3 x}{2} \text {. }
$$

Substituting (2) into (1) yields $7 x^{2}-4 k x+k^{2}-4=0$.
Since $x$ is a positive real number, this equation must have real roots.
Therefore, $\Delta=(4 k)^{2}-4 \times 7\left(k^{2}-4\right) \geqslant 0$.
Solving this, we get $k^{2} \leqslant \frac{28}{3}$. Hence, $k \leqslant \frac{2 \sqrt{21}}{3}$.
Also, $k=3 x+2 y=x+2(x+y) \geqslant x+2 \geqslant 2$, so,
$2 \leqslant k \leqslant \frac{2 \sqrt{21}}{3}$.
When $x=\frac{4 \sqrt{21}}{21}$, $k=\frac{2 \sqrt{21}}{3}$;
When point $P$ coincides with point $A$, $k=2$.
Therefore, the maximum value sought is $\frac{2 \sqrt{21}}{3}$, and the minimum value is 2.