解???$：(1)$???由二次函數(shù)圖象的頂點(diǎn)為???$(1，3)，$???
可設(shè)二次函數(shù)解析式為???$y=a{(x-1)}^2+3，$???
將???$x=3，$??????$y=-1$???代入可得???$-1=a×{(3-1)}^2+3$???
解得???$a=-1，$???
所以二次函數(shù)解析式為：???$y=-{(x-1)}^2+3=-{x}^2+2x+2；$???
將???$x=3，$??????$y=-1$???代入一次函數(shù)解析式???$y=x+m$???可得:
???$-1=3+m$???
得到???$m=-4$???
所以一次函數(shù)解析式為???$：y=x-4$???
???$(2)$???聯(lián)立兩個(gè)解析式可得方程組:
???$\{\begin{array}{l}y=-{x}^2+2x+2\\y=x-4\end{array}.$???
則???$x-4=-{x}^2+2x+2$???
化簡(jiǎn)得：???$(x-3)(x+2)=0$???
???$∴{x}_1=3，$??????${x}_2=-2$???
當(dāng)???${x}_1=3$???時(shí)，???${y}_1=3-4=-1$???
當(dāng)???${x}_2=-2$???時(shí)，???${y}_1=-2-4=-6$???
方程組的解為???$\{\begin{array}{l}{x}_1=3\\{y}_1=-1\end{array}，$??????$\{\begin{array}{l}{x}_2=-2\\{y}_2=-6\end{array}.$???
所以另一個(gè)交點(diǎn)坐標(biāo)為???$(-2，-6)$??
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