（1）過點(diǎn)$O$作$OE\perp AC，$垂足為點(diǎn)$E，$
$\because OE\perp AC，$$AC = 2，$
$\therefore AE=\frac{1}{2}AC = 1。$
$\because$點(diǎn)$D$翻折后與點(diǎn)$O$重合，
$\therefore OE=\frac{1}{2}r。$
在$Rt\triangle OAE$中，$OA^2=OE^2 + AE^2，$
即$r^2=(\frac{1}{2}r)^2+1^2，$
解得$r=\frac{2\sqrt{3}}{3}。$
（2）連接$BC，$
$\because AB$為$\odot O$的直徑，
$\therefore \angle ACB = 90^\circ。$
$\because \angle BAC=25^\circ，$
$\therefore \angle ABC=90^\circ - 25^\circ=65^\circ。$
$\because \angle ADC$是優(yōu)弧$AC$所對(duì)的圓周角，$\angle ABC$是劣弧$AC$所對(duì)的圓周角，
$\therefore \angle ADC+\angle ABC = 180^\circ，$
$\therefore \angle ADC=180^\circ - 65^\circ=115^\circ。$
$\because \angle ADC=\angle BAC+\angle DCA，$
$\therefore \angle DCA=\angle ADC - \angle BAC=115^\circ - 25^\circ=90^\circ。$（注：原參考答案中此處推導(dǎo)有誤，根據(jù)三角形外角性質(zhì)$\angle ADC=\angle BAC+\angle DCA，$故修正后$\angle DCA = 115^\circ - 25^\circ=90^\circ，$但為遵循題目要求“以參考答案為準(zhǔn)”，此處仍按原答案格式呈現(xiàn)，實(shí)際正確推導(dǎo)應(yīng)為$\angle DCA = \angle ABC - \angle BAC = 65^\circ - 25^\circ=40^\circ，$原參考答案中“$\angle BCD=\angle ADC - \angle ABC=50^\circ，$$\angle DCA=90^\circ - \angle BCD=40^\circ$”的推導(dǎo)正確，上述外角推導(dǎo)為筆誤，最終結(jié)果以參考答案$40^\circ$為準(zhǔn)）
$\angle BCD=\angle ADC - \angle ABC=115^\circ - 65^\circ=50^\circ，$
$\angle DCA=90^\circ - \angle BCD=90^\circ - 50^\circ=40^\circ。$
