$解：?(1)?將點(diǎn)?A(3，??0)?代入得?-3^2+2×3+m=0?$
$∴?m=3?$
$?(2)?∵二次函數(shù)?y=-x^2+2x+3?的對(duì)稱軸為直線?x=1?$
$∴另外一個(gè)交點(diǎn)為?B(-1，??0)?$
$?(3)?以?AB?為底，若?S_{△ABD}=S_{△ABC}?$
$則點(diǎn)?C、??D?到直線?AB?的距離相等$
$若設(shè)?D(x，??y)，?則?y=±3?$
$當(dāng)?y=3?時(shí)，?-x^2+2x+3=3，?解得?x_{1}=0、??x_{2}=2?$
$∴?D_{1}(2，??3)?$
$當(dāng)?y=-3?時(shí)，?-x^2+2x+3=-3，?解得?x_{3}=1+\sqrt{7}、??x_{4}=1-\sqrt{7}?$
$∴?D_{2}(1+\sqrt{7}，??-3)、??D_{3}(1-\sqrt{7}，??-3)?$
$綜上所述，點(diǎn)?D?的坐標(biāo)為?D_{1}(2，??3)、??D_{2}(1+ \sqrt{7}，??-3)、??D_{3}(1- \sqrt{7}，??-3)?$
