[Solution] Let the arc lengths corresponding to the chords of lengths 3 and 5 be $2 \alpha$ and $2 \beta$, respectively. Connect $A C$,
we know $\angle A M C=\frac{1}{2} \cdot \frac{2}{3} \cdot 360^{\circ}=120^{\circ}$
In $\triangle A M C$, by the cosine rule, we can find $A C=7$. Then, by the sine rule,
$$
\frac{3}{\sin \alpha}=\frac{5}{\sin \beta}=\frac{7}{\sin 120^{\circ}}=\frac{7}{\sqrt{3} / 2},
$$

we get $\sin \alpha=\frac{3}{14} \sqrt{3}, \sin \beta=\frac{5}{14} \sqrt{3}$, and $\sin \alpha: \sin \beta=3: 5$. Connect $C N$, in $\triangle C M N$, $\frac{M N}{\sin \angle M C N}=\frac{M C}{\sin \angle M N C}$. Then
$$
\begin{aligned}
M N & =\frac{5 \sin 3 \alpha}{\sin \beta}=\frac{5\left(3 \sin \alpha-4 \sin ^{3} \alpha\right)}{\sin \beta} \\
& =5\left(3 \cdot \frac{\sin \alpha}{\sin \beta}-4 \cdot \frac{\sin \alpha}{\sin \beta} \cdot \sin ^{2} \alpha\right) \\
& =9-12 \sin ^{2} \alpha=9-12\left(\frac{3}{14} \sqrt{3}\right)^{2}=\frac{360}{49} .
\end{aligned}
$$

Thus, $m+n=360+49=409$.
Therefore, the answer is $(E)$.
$20 \cdot 99$ A cyclic hexagon has three consecutive sides of length 3 each, and the other three sides of length 5 each. A chord of the circle divides it into two quadrilaterals: one quadrilateral has three sides of length 3 each, and the other quadrilateral has three sides of length 5 each. If the length of the chord is $\frac{m}{n}$, where $m, n$ are relatively prime positive integers, then $m+n$ equals
(A) 309 .
(B) 349 .
(C) 369 .
(D) 389 .
(E) 409 .
(47th American High School Mathematics Examination, 1996)