$解：（1）\because BE\parallel CD，\therefore \angle BEC=\angle DCE.$
$\because \angle CEB=\angle CED，\therefore \angle CED=\angle DCE.$
$\therefore CD=DE=25，故道路CD的长为25 m\ $
$（3）由（2）可得EB垂直平分AC，$
$根据两点之间线段最短可得AD，EB的交点G到A，E，D，B的距离之和最小，$
$又GA=GC，则到四栋距离最小的点即为点G，如图所示\ $
$（4）\because DC\parallel EB，EB\perp AC，\therefore AC\perp DC，\therefore \angle ADC=90^\circ.$
$\because点G在EB上，即AC的垂直平分线上，$
$\therefore GA=GC，\therefore \angle GAC=\angle GCA.$
$\because \angle GAC+\angle GCD=90^\circ，\angle GCA+\angle GDC=90^\circ，$
$\therefore \angle GCD=\angle GDC.\therefore GD=GC，\therefore GA=GD=GC，$
$\therefore CG+DG+EG+BG=AG+GD+EG+GB=AD+EB$
$=\sqrt{DC^2+AC^2}+EB=\sqrt{25^2+48^2}+50=(\sqrt{2929}+50)(\text{m}) $
