$解:(1)設(shè)直線(xiàn)AD表達(dá)式為y=ax+b$
$由 A(3,5),E(-2,0)$
$∴\begin{cases}{ 3a+b=5 }\ \\ {-2a+b=0\ } \end{cases}解得\begin{cases}{ a=1 }\ \\ { b=2 } \end{cases}$
$直線(xiàn)AD的函數(shù)表達(dá)式為y=x+2$
$(2)∵點(diǎn)A(3,5)關(guān)于原點(diǎn)O的對(duì)稱(chēng)點(diǎn)為點(diǎn)C$
$∴點(diǎn)C的坐標(biāo)為 (-3,-5)$
$∵CD//y軸，∴設(shè)點(diǎn)D的坐標(biāo)為(-3,a),∴a=-3+2=-1$
$∴點(diǎn)D的坐標(biāo)為(-3,1)$
$∵反比例函數(shù)y=\frac{k}{x}的圖像經(jīng)過(guò)點(diǎn)D,∴k=-3×(-1)=3.$
$(3)12$
