将一副三角板按如图放置, 则下列结论: (1)如果 $\angle 2=30^{\circ}$, 则有 $A C / / D E$;

(2) $\angle B A E+\angle C A D=180^{\circ}$; (3) 如果 $B C / / A D$, 则有 $\angle 2=30^{\circ}$; (4)如果 $\angle C A D=150^{\circ}$, 必有 $\angle 4=\angle C$. 正确的有 ( )

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解: $\because \angle 2=30^{\circ}$,

$\therefore \angle 1=60^{\circ}$,

$\therefore \angle 1=\angle E$,

(1) 正确;

$\because \angle 1+\angle 2=90^{\circ}, \angle 2+\angle 3=90^{\circ}$,

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又 $\because \angle E=60^{\circ}$, $\therefore A C \| D E$, 故

即

$\angle B A E+\angle C A D=\angle 1+\angle 2+\angle 2+\angle 3=90^{\circ}+90^{\circ}=180^{\circ}$, 故(2)正确;

$\because B C \| A D$,

$\therefore \angle 1+\angle 2+\angle 3+\angle C=180^{\circ}$,

又 $\because \angle C=45^{\circ}, \angle 1+\angle 2=90^{\circ}$,

$\therefore \angle 3=45^{\circ}$,

$\therefore \angle 2=90^{\circ}-45^{\circ}=45^{\circ}$, 故(3)错误;

$\because \angle D=30^{\circ}, \angle C A D=150^{\circ}$,

$\therefore \angle C A D+\angle D=180^{\circ}$,

$\therefore A C \| D E$,

$\therefore \angle 4=\angle C$, 故(4)正确.

故选 $A$.