$解：(1) 设AB所在直线的函数表达式为y_1 = ax + b,$
$将A(0,4),B(2,0)代入,得\begin{cases} b = 4, \\ 2k + b = 0, \end{cases}$
$解得\begin{cases} k = -2, \\ b = 4, \end{cases}$
$∴ AB所在直线的函数表达式为y_1 = -2x + 4$
$(3)∵ 点A,D,E的坐标分别为(0,4),(\frac{1}{2},0),(\frac{2}{3},\frac{8}{3}),$
$同理(1),直线DE对应的函数表达式为y_2 = 16x - 8,$
$直线AD对应的函数表达式为y_3 = -8x + 4,$
$∴ 分两种情况:$
$① y_1 = 2y_3,即-2x + 4 = 2(-8x + 4),解得x = \frac{2}{7};$
$② y_1 = 2y_2,即-2x + 4 = 2(16x - 8),解得x = \frac{10}{17}.\ $
$综上所述,当x为\frac{2}{7}或\frac{10}{17}时,小狗与乙的距离等于其与甲的距离$
