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 解：因为$\angle ACB = 90^{\circ}，$$AB = 10\mathrm{cm}，$$BC = 8\mathrm{cm}，$所以$AC = \sqrt{AB^{2}-BC^{2}} = 6\mathrm{cm}，$所以$S_{\triangle ABC} = \frac{1}{2}AC\cdot BC = 24\mathrm{cm}^{2}。$由题意，得$AP = 2t\mathrm{cm}，$$CQ = t\mathrm{cm}。$因为$8\div1 = 8(\mathrm{s})，$所以$0\leq t\leq8。$分类讨论如下：

 ① 当$0\leq t\leq3$时，点$P$在线段$AC$上，则$CP = AC - AP = (6 - 2t)\mathrm{cm}，$所以$S_{\triangle PQC} = \frac{1}{2}CP\cdot CQ = (-t^{2} + 3t)\mathrm{cm}^{2}。$当$S_{\triangle PQC} = \frac{1}{6}S_{\triangle ABC}$时，$-t^{2} + 3t = \frac{1}{6}\times24。$整理，得$t^{2} - 3t + 4 = 0，$该方程无解；

 ② 当$3\lt t\leq8$时，点$P$在线段$AC$的延长线上，则$CP = AP - AC = (2t - 6)\mathrm{cm}，$所以$S_{\triangle PQC} = \frac{1}{2}CP\cdot CQ = (t^{2} - 3t)\mathrm{cm}^{2}。$当$S_{\triangle PQC} = \frac{1}{6}S_{\triangle ABC}$时，$t^{2} - 3t = \frac{1}{6}\times24。$整理，得$t^{2} - 3t - 4 = 0，$解得$t_{1} = 4，$$t_{2} = -1$（不合题意，舍去）。

 综上所述，存在某一时刻，使得$\triangle PQC$的面积是$\triangle ABC$面积的$\frac{1}{6}，$此时$t$的值为$4。$

$432\sqrt{2}$

 解：过点$P$作$PG\perp OC$于点$G，$则$\angle OGP = 90^{\circ}。$因为$\angle AOC = 90^{\circ}，$$OP$平分$\angle AOC，$所以$\angle POG = \frac{1}{2}\angle AOC = 45^{\circ}，$所以$\triangle OPG$是等腰直角三角形，所以$OG = PG。$由题意，得$OP = \sqrt{2}t，$$OQ = 2t，$所以$Q(2t,0)。$因为$OG^{2} + PG^{2} = OP^{2}，$所以$OG = PG = t，$所以$P(t,t)。$因为$Q(2t,0)，$$B(6,2)，$所以$PB^{2} = (6 - t)^{2} + (2 - t)^{2} = 2t^{2} - 16t + 40，$$QB^{2} = (6 - 2t)^{2} + 2^{2} = 4t^{2} - 24t + 40，$$PQ^{2} = (2t - t)^{2} + (0 - t)^{2} = 2t^{2}。$分类讨论如下：

 ① 若$\angle PQB = 90^{\circ}，$则有$PQ^{2} + QB^{2} = PB^{2}，$即$2t^{2} + 4t^{2} - 24t + 40 = 2t^{2} - 16t + 40。$整理，得$t^{2} - 2t = 0，$解得$t_{1} = 2，$$t_{2} = 0$（不合题意，舍去）；

 ② 若$\angle PBQ = 90^{\circ}，$则有$PB^{2} + QB^{2} = PQ^{2}，$即$2t^{2} - 16t + 40 + 4t^{2} - 24t + 40 = 2t^{2}。$整理，得$t^{2} - 10t + 20 = 0，$解得$t_{1} = 5 + \sqrt{5}，$$t_{2} = 5 - \sqrt{5}；$

 ③ 若$\angle QPB = 90^{\circ}，$则有$PQ^{2} + PB^{2} = QB^{2}，$即$2t^{2} + 2t^{2} - 16t + 40 = 4t^{2} - 24t + 40。$整理，得$8t = 0，$解得$t = 0$（不合题意，舍去）。

 综上所述，当$t = 2$或$5 + \sqrt{5}$或$5 - \sqrt{5}$时，$\triangle PQB$为直角三角形。