解：(1) 因为四边形$ABCD$是菱形，
$AC = 8，$$BD = 6，$所以$AO = 4，$$BO = 3，$$AC\perp BD。$
当$t = 1$时，$AM = 2\times1 = 2，$$BN = 1\times1 = 1，$
则$OM = 4 - 2 = 2，$$ON = 3 - 1 = 2。$
根据勾股定理
$MN=\sqrt{OM^{2}+ON^{2}}=\sqrt{2^{2}+2^{2}}=2\sqrt{2}$（cm）。
(2) ①当$2\lt t\lt3$时，$AM = 2t，$$OM = 2t - 4。$
②$ON = 3 - t，$根据勾股定理$MN^{2}=OM^{2}+ON^{2}，$
$W=(2t - 4)^{2}+(3 - t)^{2}=4t^{2}-16t + 16+9 - 6t+t^{2}=5t^{2}-22t + 25。$
(3) $S_{\triangle MON}=\frac{1}{2}OM\cdot ON=\frac{1}{4}。$
当$0\lt t\lt2$时，$OM = 4 - 2t，$$ON = 3 - t，$
则$\frac{1}{2}(4 - 2t)(3 - t)=\frac{1}{4}，$
$2(4 - 2t)(3 - t)=1，$$2(12-4t-6t + 2t^{2})=1，$
$24-20t + 4t^{2}=1，$$4t^{2}-20t + 23 = 0，$
$t=\frac{20\pm\sqrt{400 - 368}}{8}=\frac{20\pm\sqrt{32}}{8}=\frac{20\pm4\sqrt{2}}{8}=\frac{5\pm\sqrt{2}}{2}，$
因为$0\lt t\lt2，$所以$t=\frac{5 - \sqrt{2}}{2}。$
当$2\lt t\lt3$时，$OM = 2t - 4，$$ON = 3 - t，$
则$\frac{1}{2}(2t - 4)(3 - t)=\frac{1}{4}，$
$(2t - 4)(3 - t)=\frac{1}{2}，$$6t-2t^{2}-12 + 4t=\frac{1}{2}，$
$-2t^{2}+10t-12-\frac{1}{2}=0，$$-2t^{2}+10t-\frac{25}{2}=0，$
$4t^{2}-20t + 25 = 0，$$(2t - 5)^{2}=0，$$t=\frac{5}{2}。$
当$3\lt t\lt4$时，$OM = 2t - 4，$$ON = t - 3，$
则$\frac{1}{2}(2t - 4)(t - 3)=\frac{1}{4}，$
$(2t - 4)(t - 3)=\frac{1}{2}，$$2t^{2}-6t-4t + 12=\frac{1}{2}，$$2t^{2}-10t + 12-\frac{1}{2}=0，$$4t^{2}-20t + 23 = 0，$
$t=\frac{20\pm\sqrt{400 - 368}}{8}=\frac{20\pm\sqrt{32}}{8}=\frac{20\pm4\sqrt{2}}{8}=\frac{5\pm\sqrt{2}}{2}，$
因为$3\lt t\lt4，$所以$t=\frac{5 + \sqrt{2}}{2}。$
所以$t$的值为$\frac{5 - \sqrt{2}}{2}s$或$\frac{5}{2}s$或$\frac{5 + \sqrt{2}}{2}s。$
