解：（1）设两动点运动$t$ s时，四边形$PBCQ$的面积是矩形$ABCD$面积的$\frac{4}{9}。$因为$AB = 6$cm，$AD = 2$cm，所以$S_{矩形ABCD} = AB\cdot AD = 12$cm²。因为$AP = 2t$cm，$CQ = t$cm，所以$BP = AB - AP = (6 - 2t)$cm。因为$BC = AD = 2$cm，所以$S_{四边形PBCQ} = \frac{1}{2}(CQ + BP)\cdot BC = \frac{1}{2}(t + 6 - 2t)\times2 = (6 - t)$cm²，所以$6 - t = \frac{4}{9}\times12，$$6 - t = \frac{16}{3}，$$t = 6 - \frac{16}{3} = \frac{2}{3}。$故两动点运动$\frac{2}{3}$s时，四边形$PBCQ$的面积是矩形$ABCD$面积的$\frac{4}{9}。$
（2）设两动点运动$x$ s时，点$P$与点$Q$之间的距离为$\sqrt{5}$cm。分类讨论如下：①当$0\leq x\leq3$时，点$P$在线段$AB$上，过点$P$作$PE\perp CD$于点$E，$则$PE = AD = 2$cm，$\angle PEQ = 90^{\circ}，$所以$PE^2 + EQ^2 = PQ^2。$因为$BP = (6 - 2x)$cm，$CQ = x$cm，所以$EQ = |BP - CQ| = |6 - 3x|$cm，所以$2^2 + |6 - 3x|^2 = (\sqrt{5})^2，$$4 + (6 - 3x)^2 = 5，$$(6 - 3x)^2 = 1，$$6 - 3x = \pm1。$当$6 - 3x = 1$时，$3x = 5，$解得$x_1 = \frac{5}{3}；$当$6 - 3x = -1$时，$3x = 7，$解得$x_2 = \frac{7}{3}。$②当$3\lt x\leq4$时，点$P$在线段$BC$上，$PC^2 + CQ^2 = PQ^2。$因为$PC = (8 - 2x)$cm，所以$(8 - 2x)^2 + x^2 = (\sqrt{5})^2，$$64 - 32x + 4x^2 + x^2 = 5，$$5x^2 - 32x + 59 = 0，$$\Delta = (-32)^2 - 4\times5\times59 = 1024 - 1180 = -156\lt0，$该方程无解。综上所述，存在某一时刻，点$P$与点$Q$之间的距离为$\sqrt{5}$cm，且运动所需的时间为$\frac{5}{3}$s或$\frac{7}{3}$s。
