解：$(2)$在长方形$ABCD$中，
$AB=CD=5\ \mathrm {cm}，$$AD=BC=7\ \mathrm {cm}，$
$ ①$当$PD$平分$△ ABD$的面积时，
$AP=PB=\frac {5}{2}\mathrm {cm}，$
$ $即$2t=\frac {5}{2}，$
解得$t=\frac {5}{4}；$
$ ②$当$PD$平分$△ BCD$的面积时，
$CP=PB=\frac {7}{2}\mathrm {cm}，$
$ $即$2t=\frac {7}{2}+5=\frac {17}{2}，$
解得$t=\frac {17}{4}。$
$ $所以$t=\frac {5}{4}$或$t=\frac {17}{4}。$
$ (3)①35\ \mathrm {cm}^2$
$ $当$DP'$最短时，$DP⊥ BC，$即点$P $与点$C$重合，
$ S_{四边形PBP'D}=2S_{△ BDP}=2S_{△ BDC}$
$=S_{长方形ABCD}=5×7=35(\mathrm {cm}^2)。$
$ ②t=\frac {27}{8}$或$t=\frac {15}{8}$
$ $因为$S_{四边形PBP'D}=\frac {1}{4}×35=\frac {35}{4}\mathrm {cm}^2，$
且$S_{四边形PBP'D}=2S_{△ BDP}，$
所以$S_{△ BDP}=\frac {35}{8}\mathrm {cm}^2。$
当点$P $在$AB$上时，$S_{△ BDP}=\frac {1}{2}× BP× AD$
$=\frac {1}{2}×(5-2t)×7=\frac {35}{8}，$
解得$t=\frac {15}{8}；$
当点$P $在$CB$上时，
$S_{△ BDP}=\frac {1}{2}× BP× CD=\frac {1}{2}×(2t-5)×5=\frac {35}{8}，$
解得$t=\frac {27}{8}。$
综上，$t=\frac {27}{8}$或$t=\frac {15}{8}。$
