4. 72.

As shown in Figure 8, on $AB$, take $AE=OA$, and connect $OE$. Let $OA=r$. It is easy to find that
$$
\begin{array}{l}
\angle AOB=108^{\circ} \\
\angle OAB=\angle ABO \\
=\angle COD=36^{\circ} \\
\angle AEO=72^{\circ} \\
\angle BOE=36^{\circ} \\
=\angle OBE .
\end{array}
$$

Therefore, $\triangle OAE \cong \triangle COD$
$$
\Rightarrow BE=OE=CD=2 \text{. }
$$
$$
\begin{array}{l}
\text{Also, } \triangle OBE \sim \triangle ABO \Rightarrow \frac{OB}{AB}=\frac{BE}{OB} \\
\Rightarrow OB^2=AB \cdot BE \Rightarrow r^2=2(2+r) \\
\Rightarrow r=\sqrt{5}-1 .
\end{array}
$$

Furthermore, $AB=AE+BE=\sqrt{5}+1$, thus,
$$
m=n=1 \text{. }
$$

Therefore, $108m-36n=72$.
4. As shown in Figure $2, A B, C D$ are two chords of $\odot O$, $C D=2$, $A B=m+n \sqrt{5}(m, n$ are rational numbers), the degree measures of arcs $\overparen{A B}, \overparen{C D}$ are $108^{\circ}, 36^{\circ}$ respectively. Then $108 m-36 n=$ $\qquad$