解：（2）原方程两边分别平方，得$2x + 3 = x^2，$即$x^2 - 2x - 3 = 0，$因式分解得$(x - 3)(x + 1) = 0，$则$x - 3 = 0$或$x + 1 = 0，$解得$x_1 = 3，$$x_2 = -1。$经检验，当$x = -1$时，左边$=\sqrt{2\times(-1) + 3} = 1，$右边$= -1，$左边$\neq$右边，$x = -1$不合题意，舍去。故原方程的解为$x = 3。$
（3）因为四边形$ABCD$是矩形，所以$\angle A = \angle D = 90^{\circ}，$$DC = AB = 3$m。设$AP = x$m。因为$AD = 8$m，所以$PD = AD - AP = (8 - x)$m。因为$BP = \sqrt{AP^2 + AB^2} = \sqrt{x^2 + 9}$m，$CP = \sqrt{PD^2 + DC^2} = \sqrt{(8 - x)^2 + 9}$m，所以$BP + CP = (\sqrt{x^2 + 9} + \sqrt{(8 - x)^2 + 9})$m。又$BP + CP = 10$m，所以$\sqrt{x^2 + 9} + \sqrt{(8 - x)^2 + 9} = 10，$即$\sqrt{(8 - x)^2 + 9} = 10 - \sqrt{x^2 + 9}。$两边分别平方，得$(8 - x)^2 + 9 = 100 + x^2 + 9 - 20\sqrt{x^2 + 9}，$$64 - 16x + x^2 + 9 = 100 + x^2 + 9 - 20\sqrt{x^2 + 9}，$$20\sqrt{x^2 + 9} = 4x + 46，$两边同时除以$2$得$10\sqrt{x^2 + 9} = 2x + 23，$两边分别平方，得$100(x^2 + 9) = (2x + 23)^2，$$100x^2 + 900 = 4x^2 + 92x + 529，$$96x^2 - 92x + 371 = 0，$$(x - 4)^2 = 0，$解得$x_1 = x_2 = 4。$经检验，$x = 4$是原方程的解且符合题意。故$AP$的长为$4$m。
