(1)证明:如图，连接$OC.$$\because \odot O$与$AB$相切于点$C，$$\therefore OC\perp AB.$$\because OA = OB，$$\angle AOB = 120^{\circ}，$$\therefore \angle AOC=\angle BOC=\frac{1}{2}\angle AOB = 60^{\circ}.$$\because OD = OC，$$OC = OE，$$\therefore \triangle ODC$和$\triangle OCE$都是等边三角形.$\therefore OD = OC = DC，$$OC = OE = CE.$$\therefore OD = CD = CE = OE.$$\therefore$四边形$ODCE$是菱形
(2)解:如图，连接$DE$交$OC$于点$F，$$\because$四边形$ODCE$是菱形，$\therefore OF=\frac{1}{2}OC = 1，$$DE = 2DF，$$\angle OFD = 90^{\circ}.$在$Rt\triangle ODF$中，$OD = 2，$$\therefore DF=\sqrt{OD^{2}-OF^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}.$$\therefore DE = 2DF = 2\sqrt{3}.$$\therefore$涂色部分的面积$=$扇形$ODE$的面积$-$菱形$ODCE$的面积$=\frac{120\pi\times2^{2}}{360}-\frac{1}{2}OC\cdot DE=\frac{4\pi}{3}-\frac{1}{2}\times2\times2\sqrt{3}=\frac{4\pi}{3}-2\sqrt{3}$
