$解:(2)①\because 将\triangle ACD沿AD所在直线折叠，$
$点C恰好落在边AB上的E点处， $
$\therefore \angle CAD=\angle BAD，CD=DE，$
$由(1)可知，\frac {AB}{AC}=\frac {BD}{CD}， $
$又\because AC=1，AB=2，$
$\therefore \frac {2}{1}=\frac {BD}{CD}，$
$\therefore BD=2CD，$
$\because \angle BAC=90^{\circ}，$
$\therefore BC=\sqrt{A{C}^2+B{C}^2}=\sqrt{{1}^2+{2}^2}=\sqrt{5}，$
$\therefore BD+CD=\sqrt{5}，$
$\therefore 3CD=\sqrt{5}，$
$\therefore CD=\frac {\sqrt{5}}{3}；$
$\therefore DE=\frac {\sqrt{5}}{3}.$
$②\because 将\triangle ACD沿AD所在直线折叠，$
$点C恰好落在边AB上的E点处，$
$\therefore \angle CAD=\angle BAD，CD=DE，\angle C=\angle AED=\alpha ，$
$\therefore \tan \angle C=\tan \alpha =\frac {AB}{AC}，$
$由(1)可知，\frac {AB}{AC}=\frac {BD}{CD}，$
$\therefore \tan \alpha =\frac {BD}{CD}，$
$\therefore BD=CD\cdot \tan \alpha ，$
$又\because BC=BD+CD=m，$
$\therefore CD\cdot \tan \alpha +CD=m，$
$\therefore CD=\frac {m}{1+tanα}，$
$\therefore DE=\frac {m}{1+tanα}.$
