$\ 解：（1）因为四边形ABCD为矩形$
$所以AD = BC=\sqrt{3}，CD = AB = 1，\angle B=\angle C = 90^{\circ}$
$设CE = x，则DE = CD - CE = 1 - x$
$由折叠的性质，得AB' = AB = 1，B'C' = BC=\sqrt{3}，C'E = CE = x$
$\angle B'=\angle B = 90^{\circ}，\angle C'=\angle C = 90^{\circ}$
$则B'D=\sqrt{AD^{2}-AB'^{2}}=\sqrt{3 - 1}=\sqrt{2}$
$所以C'D = B'C'-B'D=\sqrt{3}-\sqrt{2}$
$因为C'D^{2}+C'E^{2}=DE^{2}，所以(\sqrt{3}-\sqrt{2})^{2}+x^{2}=(1 - x)^{2}$
$解得x=\sqrt{6}-2$
$故CE的长为\sqrt{6}-2$
$（2）连接AC，AC'。因为AB = 1，BC=\sqrt{3}，\angle B = 90^{\circ}$
$所以AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{1 + 3}=2$
$由折叠的性质，得AC' = AC = 2，所以点C'的运动路径是以点A为圆心，2为半径的圆弧$
$当点E运动到点D时，点C'恰好在CD的延长线上，且C'D = CD = 1$
$所以CC' = C'D + CD = 2，所以CC' = AC' = AC，所以\triangle ACC'为等边三角形$
$所以\angle CAC' = 60^{\circ}$
$所以点C'运动的路径长为\frac{60\pi\times2}{180}=\frac{2\pi}{3}$
