［Solution］Obviously, $\triangle C D E$ is an equilateral triangle.
Also, $\angle 3=\angle 2, \angle D A E=\angle D E A=\angle E B C=15^{\circ}$.
Thus, $\angle 2=\angle 3=90^{\circ}+15^{\circ}=105^{\circ}$,
$$
\begin{array}{c}
\angle 1=60^{\circ}+45^{\circ}+15^{\circ}=120^{\circ}, \angle 4=45^{\circ}-15^{\circ}=30^{\circ}, \\
(\angle 1+\angle 2):(\angle 3+\angle 4)=225: 135=5: 3 .
\end{array}
$$

In summary, (1) and (3) are true, (2) is false.
Therefore, the answer is $(C)$.
18.71 As shown in the figure, if $ABCD$ is a square, $DE=EC, \angle CDE=60^{\circ}$, then among the following relationships:
$$
\begin{array}{l}
\text { (1) } \angle 1: \angle 4=4: 1 ; \\
\text { (2) } \angle 1: \angle 3=1: 1 ; \\
\text { (3) }(\angle 1+\angle 2):(\angle 3+\angle 4)=5: 3 .
\end{array}
$$

Which of the following is correct?
(A) (1), (2), (3).
(B) Only (1).
(C) Only (1) and (3).
(D) Only (2) and (3).
(China Guangzhou, Wuhan, Fuzhou, etc. five cities junior high school mathematics competition, 1989)