Given point $O$ is the center of the unit square $ABCD$, point $P$ is on side $BC$ (point $P$ does not coincide with vertices $B$, $C$), line $DP$ intersects the extension of side $AB$ at point $Q$, $DP$ intersects diagonal $AC$ at point $E$, and line $QO$ intersects side $AD$ at point $F$.
(1) Compare the size of $AF + BP$ with $AB$, and explain the reason;
(2) Compare the size of $AF + BP$ with $\frac{3}{2} AB$, and explain the reason;
(3) If $EF \parallel AB$, find the length of $CP$;
(4) If $EF \parallel OP$, find the length of $CP$.
Solution: (1) As shown in Figure 3, let $O Q$ intersect $B C$ at point $T$. Since point $P$ is on side $B C$ and point $O$ is inside $\angle P Q A$, point $T$ is inside $\angle P Q A$. Therefore,
$$
\begin{array}{l}
A F+B P \\
>A F+B T=A F+D F=A D=A B .
\end{array}
$$
(2) Let $C P=x$, then $B P=1-x$.

By the similarity of right triangles $\triangle P C D \backsim \mathrm{Rt} \triangle P B Q$, we have $\frac{B Q}{C D}=\frac{B P}{C P}$, which means
$$
B Q=\frac{C D \cdot B P}{C P}=\frac{1 \times(1-x)}{x}=\frac{1}{x}-1 .
$$

Draw $O H \perp A B$ at point $H$, then $O H=A H=H B=\frac{1}{2}$.
Since $A F / / O H$, we have $\frac{A F}{O H}=\frac{A Q}{H Q}$, which means
$A F=\frac{O H \cdot A Q}{H Q}=\frac{\frac{1}{2}\left(1+\frac{1}{x}-1\right)}{\frac{1}{2}+\frac{1}{x}-1}=\frac{1}{2-x}$.
Thus, $A F+B P=\frac{1}{2-x}+1-x<\frac{3}{2} A B$
$\Leftrightarrow 1<(2-x)\left(\frac{1}{2}+x\right) \Leftrightarrow x\left(x-\frac{3}{2}\right)<0$.
Since $0<x<1$, the above inequality holds.
Therefore, $\frac{3}{2} A B$ is longer than $A F+B P$.
(3) Draw $E G \perp A B$ at point $G$, and extend $G E$ to intersect side $C D$ at point $K$, then $\triangle E C D \backsim \triangle E A Q$. Therefore, $\frac{E K}{E Q}=\frac{C D}{A Q}$.
Thus, $\frac{E K}{E G}+1=\frac{1}{1+\frac{1}{x}-1}+1=x+1$, which means
$$
\frac{G K}{E G}=x+1 \Rightarrow E G=\frac{1}{x+1} \text {. }
$$

Since $E F / / A B$, we have $A F=E G \Rightarrow \frac{1}{2-x}=\frac{1}{x+1}$.
Solving this, we get $x=\frac{1}{2}$. Therefore, $C P=x=\frac{1}{2}$.
(4) Clearly, $A G=E G=\frac{1}{x+1}$.

Since $E F / / O P$, we have $\frac{E G-A F}{A G}=\frac{P B-O H}{H B}$, which means
$$
\frac{\frac{1}{x+1}-\frac{1}{2-x}}{\frac{1}{x+1}}=\frac{1-x-\frac{1}{2}}{\frac{1}{2}} \text {. }
$$

Simplifying and factoring, we get
$$
(2 x-1)(x-1)=0 \text {. }
$$

It is easy to see that $C P=x=\frac{1}{2}$.