Three, on a plane, there are four points $A, B, P, Q$, where $A$ and $B$ are fixed points, $A B=\sqrt{3} ; P$ and $Q$ are moving points, and they satisfy the relationship $A P=P Q=Q B=1$. Also, the areas of $\triangle A P B$ and $\triangle P Q B$ are $S_{1}, S_{2}$, respectively. Try to find:
(1) The range of $S_{\mathrm{i}}+S_{2}^{2}$;
(2) When $S_{1}^{2}+S_{2}^{2}$ is maximized, what kind of triangle is $\triangle A P B$?
Three, (1) According to the cosine theorem:
$$
P B^{2}=1+3-2 \sqrt{3} \cos A=1+1-2 \cos Q \text {. }
$$

Then $\cos Q=\sqrt{3} \cos A-1$.
(I)
Thus, $S_{1}^{2}+S_{i}^{2}=(\sqrt{3}-\sin A)^{2}+\left(\frac{1}{2} \sin Q\right)^{2}$
$$
\begin{array}{l}
=\frac{3}{4}\left(1-\cos ^{2} A\right)+\frac{1}{4}(1- \\
\cos ^{2} Q \text { ) } \\
=-\frac{3}{2}\left(\cos A-\frac{1}{2 \sqrt{3}}\right)^{2}+\frac{7}{8} . \\
\end{array}
$$

From (1) $0 \leqslant \cos A<1$. (Otherwise, if $\cos A<0$, it would lead to $\cos Q=\sqrt{3} \cos A-1<-1$, which is impossible) Therefore, when $\cos A=-\frac{1}{2 \sqrt{3}}$, $S_{1}^{2}+S_{2}^{2}$ reaches its maximum value $\frac{7}{8}$.
$$
\text { Hence } \frac{2 \sqrt{3}-3}{4} \leqslant S_{1}^{2}+S_{2}^{2} \leqslant-\frac{7}{8} \text {. }
$$
(2) When $S_{1}^{2}+S_{2}^{2}$ is maximized, $2 \sqrt{3} \cos A=1$. From (1),
$$
\begin{array}{l}
P B^{2}=1+3-1=3 . \\
\therefore P B=\sqrt{3}=A B .
\end{array}
$$

Therefore, $\triangle A P B$ is an isosceles triangle.
(Provided by Shandong Liang Weiliang)