Five. (20 points) As shown in Figure 3, let $A B C D-A_{1} B_{1} C_{1} D_{1}$ be a cube, and points $P, Q$ are on edges $A B, C_{1} D_{1}$, respectively. Let $\angle P A_{1} A=\alpha$,
$$
\begin{array}{l}
\angle Q A_{1} D_{1}=\beta, \text { and } \alpha, \beta \\
\in\left(0, \frac{\pi}{4}\right] .
\end{array}
$$
(1) Try to find the angle $\theta$ between the plane $A_{1} P Q$ and the side face $A_{1} A D D_{1}$ (expressed as an inverse trigonometric function of $\alpha, \beta$);
(2) If $\alpha+\beta=\frac{\pi}{4}$, try to find the range of $\theta$.
(1) As shown in Figure 7,  
Let $A_{1} A=1$, then
$$
\begin{array}{l}
A P=\tan \alpha, \\
D_{1} Q=\tan \beta, \\
A_{1} P=\frac{1}{\cos \alpha}, \\
A_{1} Q=\frac{1}{\cos \beta} .
\end{array}
$$
$$
\begin{array}{l}
\text { Let } \angle P A_{1} Q=\gamma \\
(0<\gamma<\pi) .
\end{array}
$$

Draw $P R \perp C_{1} D_{1}$ at $R$, and connect $A D_{1}$, then
$$
\begin{array}{l}
P R=A D_{1}=\sqrt{2}, \\
R Q=\left|D_{1} Q-A P\right|=|\tan \beta-\tan \alpha|, \\
P Q^{2}=P R^{2}+R Q^{2}=2+(\tan \beta-\tan \alpha)^{2} .
\end{array}
$$

In $\triangle A_{1} P Q$, by the cosine rule we get
$$
\begin{array}{l}
\cos \gamma=\frac{A_{1} P^{2}+A_{1} Q^{2}-P Q^{2}}{2 A_{1} P \cdot A_{1} Q} \\
=\frac{\frac{1}{\cos ^{2} \alpha}+\frac{1}{\cos ^{2} \beta}-2-\tan ^{2} \alpha-\tan ^{2} \beta+2 \tan \alpha \cdot \tan \beta}{2 \frac{1}{\cos \alpha} \cdot \frac{1}{\cos \beta}} \\
=\frac{2 \tan \alpha \cdot \tan \beta}{2 \frac{1}{\cos \alpha} \cdot \frac{1}{\cos \beta}}=\sin \alpha \cdot \sin \beta .
\end{array}
$$

Since the projection of $\triangle A_{1} P Q$ on the plane $A_{1} A D D_{1}$ is $\triangle A_{1} A D_{1}$, we have
$$
\begin{array}{l}
\cos \theta=\frac{S_{\triangle \triangle_{1} A D_{1}}}{S_{\triangle A_{1} P Q}}=\frac{\frac{1}{2} \times 1^{2}}{\frac{1}{2} A_{1} P \cdot A_{1} Q \sin \gamma} \\
=\frac{1}{\frac{1}{\cos \alpha} \cdot \frac{1}{\cos \beta} \cdot \sin \gamma}=\frac{\cos \alpha \cdot \cos \beta}{\sqrt{1-\sin ^{2} \alpha \cdot \sin ^{2} \beta}} .
\end{array}
$$

Thus, $\theta=\arccos \frac{\cos \alpha \cdot \cos \beta}{\sqrt{1-\sin ^{2} \alpha \cdot \sin ^{2} \beta}}$.
(2) Since $\alpha+\beta=\frac{\pi}{4}$, we have
$$
\begin{aligned}
\cos \theta & =\frac{\cos \alpha \cdot \cos \beta}{\sqrt{1-\left\{-\frac{1}{2}[\cos (\alpha+\beta)-\cos (\alpha-\beta)]\right\}^{2}}} \\
& =\frac{1}{2} \cdot \frac{\frac{\sqrt{2}}{2}+\cos (\alpha-\beta)}{\sqrt{1-\frac{1}{4}\left[\frac{\sqrt{2}}{2}-\cos (\alpha-\beta)\right]^{2}}} \\
& =\frac{1}{2} \cdot \frac{\frac{\sqrt{2}}{2}+\cos (\alpha-\beta)}{\sqrt{1-\frac{1}{4}\left[\cos (\alpha-\beta)-\frac{\sqrt{2}}{2}\right]^{2}}} .
\end{aligned}
$$

Since $\alpha, \beta \in\left[0, \frac{\pi}{4}\right]$, and $\alpha+\beta=\frac{\pi}{4}$, then $|\alpha-\beta| \in\left[0, \frac{\pi}{4}\right]$. Thus, $\cos (\alpha-\beta) \in\left[\frac{\sqrt{2}}{2}, 1\right]$.
Let $x=\cos (\alpha-\beta)-\frac{\sqrt{2}}{2}$, then
$$
x \in\left[0,1-\frac{\sqrt{2}}{2}\right], y=\cos \theta=\frac{1}{2} \cdot \frac{\sqrt{2}+x}{\sqrt{1-\frac{1}{4} x^{2}}} .
$$

Clearly, $\sqrt{1-\frac{1}{4} x^{2}}$ is a strictly decreasing function of $x$, and $\frac{1}{\sqrt{1-\frac{1}{4} x^{2}}}$ is a strictly increasing function of $x$, thus, $y=\cos \theta$ is a strictly increasing function of $x$.
When $x=0$, $y=\cos \theta=\frac{\sqrt{2}}{2}$.
When $x=1-\frac{\sqrt{2}}{2}$, $y=\cos \theta=\frac{1+\sqrt{2}}{\sqrt{5+2 \sqrt{2}}}$.
Therefore, the range of $y=\cos \theta$ is $\left[\frac{\sqrt{2}}{2}, \frac{1+\sqrt{2}}{\sqrt{5+2 \sqrt{2}}}\right]$.
Thus, the range of $\theta$ is $\left[\arccos \frac{1+\sqrt{2}}{\sqrt{5+2 \sqrt{2}}}, \frac{\pi}{4}\right]$.