（5分）已知直线$x-my+1=0$与$\odot C:(x-1)^{2}+y^{2}=4$交于$A$，$B$两点，写出满足“$\Delta ABC$面积为$\dfrac{8}{5}$”的$m$的一个值____．
分析：由“$\Delta ABC$面积为$\dfrac{8}{5}$，求得$\sin \angle ACB=\dfrac{4}{5}$，设$\dfrac{1}{2}\angle ACB=\theta$，得到$\cos \theta$，进而求得圆心到直线的距离，结合点到直线的距离公式，列出方程，即可求解．
解：由圆$C:(x-1)^{2}+y^{2}=4$，可得圆心坐标为$C(1,0)$，半径为$r=2$，
因为$\Delta ABC$的面积为$\dfrac{8}{5}$，可得$S_{\Delta ABC}=\dfrac{1}{2}\times 2\times 2\times \sin \angle ACB=\dfrac{8}{5}$，
解得$\sin \angle ACB=\dfrac{4}{5}$，设$\dfrac{1}{2}\angle ACB=\theta$所以$\therefore 2\sin \theta \cos \theta =\dfrac{4}{5}$，
可得$\dfrac{2\sin \theta \cos \theta }{si{n}^{2}\theta +co{s}^{2}\theta }=\dfrac{4}{5}$，$\therefore$$\dfrac{2\tan \theta }{ta{n}^{2}\theta +1}=\dfrac{4}{5}$，$\therefore \tan \theta =\dfrac{1}{2}$或$\tan \theta =2$，
$\therefore \cos \theta =\dfrac{2}{\sqrt{5}}$或$\cos \theta =\dfrac{1}{\sqrt{5}}$，
$\therefore$圆心到直线$x-my+1=0$的距离$d=\dfrac{4}{\sqrt{5}}$或$\dfrac{2}{\sqrt{5}}$，
$\therefore$$\dfrac{2}{\sqrt{1+{m}^{2}}}=\dfrac{4}{\sqrt{5}}$或$\dfrac{2}{\sqrt{1+{m}^{2}}}=\dfrac{2}{\sqrt{5}}$，
解得$m=\pm \dfrac{1}{2}$或$m=\pm 2$．
故答案为：2（或$-2$或$\dfrac{1}{2}$或$-\dfrac{1}{2})$．
点评：本题考查了直线与圆的位置关系，属于中档题．