[Solution] Extend $B A, C D$ to intersect at point $H$.
$$
\begin{array}{ll}
\because & \angle B=30^{\circ}, \angle C=60^{\circ}, \\
\therefore & \angle B H C=90^{\circ} .
\end{array}
$$

In the right triangle $\triangle B H C$,
$\because \quad M, N$ are the midpoints of $B C, A D$ respectively,
$\therefore M, N, H$ are collinear,
then $H M=\frac{1}{2} B C=\frac{7}{2}$, thus $H N=\frac{1}{2}$, so $A N=H N=\frac{1}{2}$, hence $A D=1, E F=\frac{1}{2}(A D+B C)=4$.
Therefore, the answer is $(A)$.
18.13 In trapezoid $A B C D$, $A D / / B C, \angle B=30^{\circ}, \angle C=60^{\circ}, E$, $M$, $F$, $N$ are the midpoints of $A B$, $B C$, $C D$, $D A$ respectively. Given that $B C=7, M N=3$. Then the value of $E F$ is
(A) 4.
(B) $4 \frac{1}{2}$.
(C) 5.
(D) 6.
(China Junior High School Mathematics League, 1997)