解：$(1)$设$l_1$的函数表达式为$y=k_1x+b(k≠0)$
将点$A(0，$$2)$和$C(6，$$-2)$代入
得$\begin{cases}{b=2}\\{6k_1+b=-2}\end{cases}，$解得$\begin{cases}{k_1=-\dfrac {2}{3}}\\{b=2}\end{cases}$
∴$l_1$的表达式为$y=-\frac {2}{3}x+2$
$(2)①$当$k=\frac {4}{9}$时，
$l_2$的表达式为$y=\frac {4}{9}x+\frac {8}{9}$
联立得$\begin{cases}{y=-\dfrac {2}{3}x+2}\\{y=\dfrac {4}{9}x+\dfrac {8}{9}}\end{cases}，$解得$\begin{cases}{x=1}\\{y=\dfrac {4}{3}}\end{cases}$
则交点$M(1，$$\frac {4}{3})$
②当$y=2$时，有$2=\frac {4}{9}x+\frac {8}{9}$
解得：$x=\frac {5}{2}$
∴$P(\frac {5}{2}，$$2)$
∴点$M$到直线$AP$的距离是$2-\frac {4}{3}=\frac {2}{3}$
∴$S_{△APM}=\frac {1}{2}×\frac {5}{2}×\frac {2}{3}=\frac {5}{6}$
