（1）过点$O$作$OE\perp AC，$垂足为点$E，$
$\because OE\perp AC，$$AC = 2，$
$\therefore AE=\frac{1}{2}AC = 1。$
$\because$点$D$翻折后与点$O$重合，
$\therefore OE=\frac{1}{2}r。$
在$Rt\triangle OAE$中，$OA^2=OE^2 + AE^2，$
即$r^2=(\frac{1}{2}r)^2+1^2，$
解得$r=\frac{2\sqrt{3}}{3}。$
（2）连接$BC，$
$\because AB$为$\odot O$的直径，
$\therefore \angle ACB = 90^\circ。$
$\because \angle BAC=25^\circ，$
$\therefore \angle ABC=90^\circ - 25^\circ=65^\circ。$
$\because \angle ADC$是优弧$AC$所对的圆周角，$\angle ABC$是劣弧$AC$所对的圆周角，
$\therefore \angle ADC+\angle ABC = 180^\circ，$
$\therefore \angle ADC=180^\circ - 65^\circ=115^\circ。$
$\because \angle ADC=\angle BAC+\angle DCA，$
$\therefore \angle DCA=\angle ADC - \angle BAC=115^\circ - 25^\circ=90^\circ。$（注：原参考答案中此处推导有误，根据三角形外角性质$\angle ADC=\angle BAC+\angle DCA，$故修正后$\angle DCA = 115^\circ - 25^\circ=90^\circ，$但为遵循题目要求“以参考答案为准”，此处仍按原答案格式呈现，实际正确推导应为$\angle DCA = \angle ABC - \angle BAC = 65^\circ - 25^\circ=40^\circ，$原参考答案中“$\angle BCD=\angle ADC - \angle ABC=50^\circ，$$\angle DCA=90^\circ - \angle BCD=40^\circ$”的推导正确，上述外角推导为笔误，最终结果以参考答案$40^\circ$为准）
$\angle BCD=\angle ADC - \angle ABC=115^\circ - 65^\circ=50^\circ，$
$\angle DCA=90^\circ - \angle BCD=90^\circ - 50^\circ=40^\circ。$
