One, (50 points) As shown in Figure $3, \odot O$ is the circumcircle of the square $ABCD$, point $P$ is on the minor arc $\overparen{AB}$ ($P$ does not coincide with $A$ or $B$), $DP$ intersects $AO$ and $AB$ at points $Q$ and $T$ respectively, the tangent to $\odot O$ at point $P$ intersects the extension of $DA$ at point $E$, and the midpoint of the minor arc $\overparen{BC}$ is $F$.
(1) When are the points $F$, $T$, and $E$ collinear? Please explain your reasoning.
(2) Find the range of the ratio $\frac{PE}{PQ}$.
(1) As shown in Figure 5, let the side length of the square $ABCD$ be 1, and denote $\angle ADP = \theta (0^{\circ} < \theta < 45^{\circ})$. Since $A G > A E$, it follows that $G$ and $E$ do not coincide, meaning $F, T, E$ cannot be collinear.
$$
A G > A E
$$
$$
\Rightarrow \frac{\sin \theta}{(1+\sqrt{2}) \cos \theta - 2 \sin \theta} > \frac{\sqrt{2} \sin^2 \theta}{\sin (2 \theta + 45^{\circ})}
$$
$$
\Rightarrow \sin (2 \theta + 45^{\circ}) > \sqrt{2} \sin \theta [(1+\sqrt{2}) \cos \theta - 2 \sin \theta]
$$
(From the value of $\theta$, we know $\sin \theta, \sin (2 \theta + 45^{\circ}) > 0$)
$$
\Rightarrow \frac{1}{\sqrt{2}} (\sin 2 \theta + \cos 2 \theta)
$$
$$
> \sqrt{2} \sin \theta [(1+\sqrt{2}) \cos \theta - 2 \sin \theta]
$$
$$
\Rightarrow 2 \sin \theta \cdot \cos \theta + 1 - 2 \sin^2 \theta
$$
$$
> 2 (1+\sqrt{2}) \sin \theta \cdot \cos \theta - 4 \sin^2 \theta
$$
$$
\Rightarrow 1 + 2 \sin^2 \theta - 2 \sqrt{2} \sin \theta \cdot \cos \theta > 0
$$
$$
\Rightarrow (\cos \theta - \sqrt{2} \sin \theta)^2 + \sin^2 \theta > 0.
$$
The above inequality is clearly true.
Therefore, $F, T, E$ cannot be collinear.
(2) By the Law of Sines, we have
$$
\begin{array}{l}
\frac{P E}{\sin (45^{\circ} + \theta)} = \frac{A P}{\sin (45^{\circ} + 2 \theta)}, \\
P E = \sqrt{2} \sin \theta \frac{\sin (45^{\circ} + \theta)}{\sin (45^{\circ} + 2 \theta)}, \\
\frac{P Q}{\sin 2 \theta} = \frac{O P}{\sin (45^{\circ} + \theta)},
\end{array}
$$
and
$$
\frac{P Q}{\sin 2 \theta} = \frac{O P}{\sin (45^{\circ} + \theta)},
$$
which implies
$$
P Q = \frac{\sqrt{2}}{2} \cdot \frac{\sin 2 \theta}{\sin (45^{\circ} + \theta)}.
$$
Thus,
$$
\begin{array}{l}
\frac{P E}{P Q} = \sqrt{2} \sin \theta \frac{\sin (45^{\circ} + \theta)}{\sin (45^{\circ} + 2 \theta)} \sqrt{2} \frac{\sin (45^{\circ} + \theta)}{\sin 2 \theta} \\
= \frac{\sin^2 (45^{\circ} + \theta)}{\cos \theta \cdot \sin (45^{\circ} + 2 \theta)} \\
= \frac{\sin^2 (45^{\circ} + \theta)}{\cos \theta [\sin (45^{\circ} + \theta) \cdot \cos \theta + \cos (45^{\circ} + \theta) \cdot \sin \theta]} \\
= \frac{\sin (45^{\circ} + \theta)}{\cos \theta [\cos \theta + \frac{\sin \theta \cdot \cos (45^{\circ} + \theta)}{\sin (45^{\circ} + \theta)}]} \\
= \frac{\frac{\sqrt{2}}{2} (\sin \theta + \cos \theta)}{\cos \theta [\cos \theta + \frac{\sin \theta}{\tan (45^{\circ} + \theta)}]} \\
= \frac{1}{\sqrt{2} \cos \theta} \cdot \frac{1 + \tan \theta}{1 + \frac{\tan \theta (1 - \tan \theta)}{1 + \tan \theta}} \\
= \frac{1}{\sqrt{2} \cos \theta} \cdot \frac{(1 + \tan \theta)^2}{- \tan 2 + 2 \tan \theta + 1} \\
= \frac{1}{\sqrt{2} \cos \theta} \left(1 + \frac{2 \tan \theta}{- \tan^2 \theta + 2 \tan \theta + 1}\right) \\
= \frac{1}{\sqrt{2} \cos \theta} \left(1 + \frac{2}{-1 + 2 \cot \theta + \cot^2 \theta}\right).
\end{array}
$$
Clearly, $\frac{1}{\sqrt{2} \cos \theta}$ and $1 + \frac{2}{-1 + 2 \cot \theta + \cot^2 \theta}$ are both strictly increasing functions greater than 0. Therefore, $\frac{P E}{P Q}$ is a strictly increasing function greater than 0 with respect to the variable $\theta (0^{\circ} < \theta < 45^{\circ})$.
Thus, the range of $\frac{P E}{P Q}$ is $\left(\frac{\sqrt{2}}{2}, 2\right)$.