$解???：(1)???設(shè)???OD=x，???則???AD=CD=8-x???$
$在???Rt\triangle OCD???中，由勾股定理得$
$???C{D}^2=O{D}^2+O{C}^2，???即???{(8-x)}^2={x}^2+{4}^2???$
$解得???x=3???$
$???∴OD=3???$
$???∴D(3，0)???$
$???∵OA=8，??????OC=4???$
$???∴B(8，-4)，??????C(0，-4)???且拋物線過???B，??????C???點$
$∴拋物線的對稱軸為???x=4???$
$???∵D(3，0)???$
$∴另一個交點???E(5，0).???$
$???(2)???不存在這樣的點???P，???理由如下：$
$???{S}_{矩形OABC}=OA·OB=32???$
$若存在這樣的點???P，???設(shè)點???P???到???BC???的距離為???h???$
$則???{S}_{△PBC}=\frac {1}{2}BC·h=32???$
$???∴h=8???$
$設(shè)拋物線的解析式為???y=a{x}^2+bx+c???$
$∵拋物線過???B(8，-4)，??????C(0，-4)，??????D(3，0)???$
$???∴\{\begin{array}{l}-4=64a+8b+c\\-4=c\\0=9a+3b+c\end{array}.???$
$解得???\{\begin{array}{l}a=-\frac {4}{15}\\b=\frac {32}{15}\\c=-4\end{array}.???$
$∴拋物線解析式為???y=-\frac {4}{15}{x}^2+\frac {32}{15}x-4=-\frac {4}{15}{(x-4)}^2+\frac {4}{15}???$
$???∴p???拋物線的頂點為???(4，\frac {4}{15})???$
$∴頂點到???BC???的距離為???4+\frac {4}{15}=\frac {64}{15}\lt 8???$
$∴不存在這樣的點???P，???使???\triangle PBC???的面積等于矩形???OABC???的面積.$
