解：（2）因为一元二次方程$2x^2 - 3x - 1 = 0$的两个根分别为$m，$$n，$所以$m + n = \frac{3}{2}，$$mn = -\frac{1}{2}，$所以$\frac{n}{m} + \frac{m}{n} = \frac{n^2 + m^2}{mn} = \frac{(m + n)^2 - 2mn}{mn} = \frac{(\frac{3}{2})^2 - 2\times(-\frac{1}{2})}{-\frac{1}{2}} = \frac{\frac{9}{4} + 1}{-\frac{1}{2}} = \frac{\frac{13}{4}}{-\frac{1}{2}} = -\frac{13}{2}。$
（3）因为$2s^2 - 3s - 1 = 0，$$2t^2 - 3t - 1 = 0，$且$s\neq t，$所以可将$s，$$t$看作方程$2x^2 - 3x - 1 = 0$的两个实数根，则由一元二次方程的根与系数的关系，得$s + t = \frac{3}{2}，$$st = -\frac{1}{2}，$所以$(t - s)^2 = (s + t)^2 - 4st = (\frac{3}{2})^2 - 4\times(-\frac{1}{2}) = \frac{9}{4} + 2 = \frac{17}{4}，$所以$t - s = \pm\frac{\sqrt{17}}{2}，$所以$\frac{1}{s} - \frac{1}{t} = \frac{t - s}{st} = \pm\sqrt{17}。$
