$解：(1) 設(shè)AB所在直線的函數(shù)表達(dá)式為y_1 = ax + b,$
$將A(0,4),B(2,0)代入,得\begin{cases} b = 4, \\ 2k + b = 0, \end{cases}$
$解得\begin{cases} k = -2, \\ b = 4, \end{cases}$
$∴ AB所在直線的函數(shù)表達(dá)式為y_1 = -2x + 4$
$(3)∵ 點A,D,E的坐標(biāo)分別為(0,4),(\frac{1}{2},0),(\frac{2}{3},\frac{8}{3}),$
$同理(1),直線DE對應(yīng)的函數(shù)表達(dá)式為y_2 = 16x - 8,$
$直線AD對應(yīng)的函數(shù)表達(dá)式為y_3 = -8x + 4,$
$∴ 分兩種情況:$
$① y_1 = 2y_3,即-2x + 4 = 2(-8x + 4),解得x = \frac{2}{7};$
$② y_1 = 2y_2,即-2x + 4 = 2(16x - 8),解得x = \frac{10}{17}.\ $
$綜上所述,當(dāng)x為\frac{2}{7}或\frac{10}{17}時,小狗與乙的距離等于其與甲的距離$
