答案：22. (1)在$Rt\triangle OAB$中，$\angle OAB = 90°$，$\angle AOB = 30°$，$D$是$OB$的中点，$\therefore DO = DA=\frac{1}{2}OB$，$\therefore \angle DOA=\angle DAO = 30°$。$\because \triangle OBC$是等边三角形，$\therefore \angle COB=\angle C = 60°$，$\therefore \angle COA=\angle COB+\angle DOA = 90°$，$\therefore \angle AEO=180° - \angle COA - \angle DAO = 60°$，$\angle COA+\angle OAB = 180°$，$\therefore \angle AEO=\angle C$，$OC// AB$，$\therefore AE// BC$，$\therefore$四边形$ABCE$是平行四边形。(2)$\because \triangle OBC$为等边三角形，$\therefore OC = OB = 8$。设$OG = x$。由折叠，可知$AG = CG = 8 - x$。在$Rt\triangle OAB$中，$\because \angle OAB = 90°$，$\angle AOB = 30°$，$OB = 8$，$\therefore AB = 4$，$\therefore OA=\sqrt{OB^{2}-AB^{2}} = 4\sqrt{3}$。由(1)，知$\angle COA = 90°$，$\therefore$在$Rt\triangle OAG$中，由勾股定理，得$OG^{2}+OA^{2}=AG^{2}$，即$x^{2}+(4\sqrt{3})^{2}=(8 - x)^{2}$，解得$x = 1$，$\therefore OG$的长为$1$。