$解:連接OP_{1},OP_{2}$
$∵點(diǎn)P關(guān)于直線l,m的對稱點(diǎn)分別是點(diǎn)P_{1},P_{2},∴OP_{1}=OP=2.8,OP_{2}=OP=2.8$
$當(dāng)點(diǎn)P_{1}與P_{2}不在同一直線上時,根據(jù)三角形的三邊關(guān)系可知,P_{1}P_{2}＜2.8+2.8=5.6$
$當(dāng)點(diǎn)P_{1},P_{2}在同一直線上時,P_{1}P_{2}=2.8+2.8=5.6$
$綜上，d≤5.6$
$∴d的最大整數(shù)值為5$
