$解:(1)由題意得3a?14=0,解得a=\frac{14}{3}$ $∴點(diǎn)P的坐標(biāo)為(0,\frac{8}{3})$ $(2)∵點(diǎn)P(3a?14,2?a)位于第三象限$ $∴3a?14＜0,2?a＜0,解得2＜a＜\frac{14}{3}$ $∵點(diǎn)P的橫、縱坐標(biāo)都是整數(shù),∴a=3或a=4$ $當(dāng)a=3時,點(diǎn)P(?5,?1)$ $當(dāng)a=4時,點(diǎn)P(?2,?2)$ $∴點(diǎn)P的坐標(biāo)為(?5,?1)或(?2,?2)$ $解:依題意可知，折痕AD所在直線是四邊形OAED的對稱軸$ $∴在 Rt△ABE中,AE=AO=10,AB=8$ $BE= \sqrt{AE2?AB2}=6,∴CE=4,∴E(4,8)$ $在Rt△DCE中,DC2+CE2=DE2,又∵DE=OD$ $∴(8?OD)2+42=OD2,∴OD=5,∴D(0,5).$ $綜上,點(diǎn)D的坐標(biāo)為(0,5),點(diǎn)E的坐標(biāo)為(4,8)$
$解:(2)∵S_{△POA}=S_{△PBC},∴點(diǎn)P在對稱軸l上$ $設(shè)P(3,m).∵S_{△PAB}= S_{△POC}$ $∴\frac{1}{2}×2×(4?m)=\frac{1}{2}×6×m,∴m=1$ $∴P(3,1)$ $(3)(3,-2)$
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