$解:(1)设直线AD表达式为y=ax+b$ $由 A(3,5),E(-2,0)$ $∴\begin{cases}{ 3a+b=5 }\ \\ {-2a+b=0\ } \end{cases}解得\begin{cases}{ a=1 }\ \\ { b=2 } \end{cases}$ $直线AD的函数表达式为y=x+2$ $(2)∵点A(3,5)关于原点O的对称点为点C$ $∴点C的坐标为 (-3,-5)$ $∵CD//y轴，∴设点D的坐标为(-3,a),∴a=-3+2=-1$ $∴点D的坐标为(-3,1)$ $∵反比例函数y=\frac{k}{x}的图像经过点D,∴k=-3×(-1)=3.$ $(3)12$ $解:∵x_{1}+x_{2}=0,∴x_{1}=-x_{2}$ $∴A(x_{1},y_{2})、B(x_{2},y_{2})关于点(0,1)对称$ $∴y_{1}+y_{2}=2,∴y_{1}+y_{2}+3=5$
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