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