解：?$(1) $?當(dāng)?$y=0$?時(shí)，?$-x^2+2x+3=0$?
解得?$x_{1}=-1，$??$x_{2}=3$?
∵點(diǎn)?$A$?在點(diǎn)?$B$?的左側(cè)
∴點(diǎn)?$A、$??$B$?的坐標(biāo)分別為?$(-1，$??$0)、$??$(3，$??$0)$?
當(dāng)?$x=0$?時(shí)，?$y=3$?
∴點(diǎn)?$C$?的坐標(biāo)為?$(0，$??$3)$?
設(shè)直線?$AC$?相應(yīng)的函數(shù)表達(dá)式為?$y=k_{1}x+b_{1}(k_{1}≠0)$?
由?$A、$??$C$?兩點(diǎn)的坐標(biāo)，可求得直線?$AC$?相應(yīng)的函數(shù)表達(dá)式為?$y= 3x+3$?
∵?$y=-x^2+2x+3=-(x-1)^2+4$?
∴頂點(diǎn)?$D$?的坐標(biāo)為?$(1，$??$4) $?
?$(2) $?圖像上有三個(gè)這樣的點(diǎn)?$Q$?
如圖，①當(dāng)點(diǎn)?$Q $?在點(diǎn)?$Q 1$?位置時(shí)，點(diǎn)?$Q 1$?的縱坐標(biāo)為?$3，$?
代入二次函數(shù)表達(dá)式，可得點(diǎn)?$Q_{1} $?的坐標(biāo)為?$(2，$??$3)；$?
②當(dāng)點(diǎn)?$Q $?在點(diǎn)?$Q_{2}$?位置時(shí)，點(diǎn)?$Q 2$?的縱坐標(biāo)為?$-3，$?
代入二次函數(shù)表達(dá)式，可得點(diǎn)?$Q 2$?的坐標(biāo)為?$(1+ \sqrt{7}，$??$-3)；$?
③當(dāng)點(diǎn)?$Q $?在點(diǎn)?$Q_{3} $?位置時(shí)，點(diǎn)?$Q_{3} $?的縱坐標(biāo)為?$-3，$?
代入二次函數(shù)表達(dá)式，可得點(diǎn)?$Q_{3} $?的坐標(biāo)為?$(1-\sqrt 7，$??$-3)$?
∴滿足題意的點(diǎn)?$Q $?有：?$(2，$??$3)、$??$(1+ \sqrt{7}，$??$-3)、$??$(1 -\sqrt{7}，$??$-3) $?
