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$2\sqrt{2}$

 解：（1）如图，连接$BD。$

 $\because \angle ACD = 30^{\circ}，$$\therefore \angle B=\angle ACD = 30^{\circ}。$

 $\because AB$是$\odot O$的直径，$\therefore \angle ADB = 90^{\circ}。$

 $\therefore \angle DAB = 90^{\circ}-\angle B = 60^{\circ}。$

 （2）$\because \angle ADB = 90^{\circ}，$$\angle B = 30^{\circ}，$$AB = 4，$

 $\therefore AD=\frac{1}{2}AB = 2。$

 $\because \angle DAB = 60^{\circ}，$$DE\perp AB，$且$AB$是$\odot O$的直径，

 $\therefore \angle ADE = 30^{\circ}，$$DE = EF。$

 $\therefore AE=\frac{1}{2}AD = 1。$

 $\therefore DE=\sqrt{AD^{2}-AE^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}。$

 $\therefore DF = DE + EF = 2DE = 2\sqrt{3}。$

 解：（1）$\because \angle ACB=\frac{1}{2}\angle AOB，$$\angle BAC=\frac{1}{2}\angle BOC，$$\angle ACB = 2\angle BAC，$

 $\therefore \angle AOB = 2\angle BOC。$

 （2）如图，过点$O$作半径$OD\perp AB$于点$E，$连接$DB。$

 $\therefore AE = BE。$

 $\because \angle AOB = 2\angle BOC，$$\angle DOB=\frac{1}{2}\angle AOB，$

 $\therefore \angle DOB=\angle BOC。$

 $\therefore BD = BC。$

 $\because AB = 4，$$BC=\sqrt{5}，$$\therefore BE = 2，$$DB=\sqrt{5}。$

 在$Rt\triangle BDE$中，$\angle DEB = 90^{\circ}，$

 $\therefore DE=\sqrt{BD^{2}-BE^{2}}=\sqrt{(\sqrt{5})^{2}-2^{2}} = 1。$

 在$Rt\triangle BOE$中，$\angle OEB = 90^{\circ}，$$OB^{2}=(OB - 1)^{2}+2^{2}，$

 $OB^{2}=OB^{2}-2OB + 1+4，$

 $2OB = 5，$

 解得$OB=\frac{5}{2}，$即$\odot O$的半径是$\frac{5}{2}。$

 解：（1）$\because \angle ADC=\angle BCD = 90^{\circ}，$

 $\therefore AC，$$BD$是$\odot O$的直径，且交点为圆心$O。$

 $\because AD = CD，$$AO = CO，$

 $\therefore AC\perp BD。$

 （2）连接$CO$并延长，交$\odot O$于点$K，$连接$DK，$$BC，$则$\angle KDC = 90^{\circ}。$

 $\therefore \angle K+\angle KCD = 90^{\circ}。$

 $\because AC\perp BD，$$\therefore \angle ACB+\angle EBC = 90^{\circ}。$

 $\because \angle K=\angle EBC，$$\therefore \angle KCD=\angle ACB。$

 $\therefore \overset{\frown}{DK}=\overset{\frown}{AB}。$

 $\therefore DK = AB = 2。$

 在$Rt\triangle KCD$中，$\because DC = 4，$$DK = 2，$

 $\therefore KC=\sqrt{2^{2}+4^{2}}=2\sqrt{5}。$

 $\therefore \odot O$的半径为$\sqrt{5}。$