（1）证明：因为$\angle COD = \angle AOB = 90^{\circ}，$所以$\angle AOC + \angle AOD = \angle BOD + \angle AOD，$所以$\angle AOC = \angle BOD。$
在$\triangle AOC$和$\triangle BOD$中，$\begin{cases}AO = BO\\\angle AOC = \angle BOD\\CO = DO\end{cases}，$所以$\triangle AOC\cong\triangle BOD，$所以$AC = BD。$
（2）解：如图，以点$O$为圆心，$OC$为半径画弧，分别交$OA，$$OB$于点$F，$$E。$
因为$\angle COF = \angle EOD，$所以$S_{扇形OCF} = S_{扇形OED}。$
因为$\triangle AOC\cong\triangle BOD，$所以$S_{\triangle AOC} = S_{\triangle BOD}，$所以$S_{\triangle AOC} - S_{扇形OCF} = S_{\triangle BOD} - S_{扇形OED}，$即$S = S'，$所以$S_{阴影} = S_{扇形OAB} - S_{扇形OEF}。$
设$OA = r，$则$S_{扇形OAB} = \frac{90\pi r^{2}}{360}=\frac{\pi r^{2}}{4}。$
因为$OC = 3，$所以$S_{扇形OEF} = \frac{90\pi\times3^{2}}{360}=\frac{9\pi}{4}。$
又$S_{阴影}=\frac{7\pi}{4}，$所以$\frac{\pi r^{2}}{4}-\frac{9\pi}{4}=\frac{7\pi}{4}，$
等式两边同时乘以$4$得：$\pi r^{2}-9\pi = 7\pi，$
移项得：$\pi r^{2}=7\pi + 9\pi，$
即$\pi r^{2}=16\pi，$
等式两边同时除以$\pi$得：$r^{2}=16，$
解得$r = 4$（负值舍去）。
故$OA$的长为$4。$
