$解:(1)由\begin{cases}{ y=-\frac {1}{2}x-1 }\ \\ { y=-2x+2 } \end{cases}解得\begin{cases}{\ x=2}\ \\ { y=-2 } \end{cases}$
$∴P(2,-2)$
$(2)把y=0分別代入兩直線有$
$-\frac {1}{2}x-1=0,解得x=-2\ ∴A(-2,0)$
$-2x+2=0,解得x=1\ \ ∴B(1,0)$
$S_{△PAB}=\frac {1}{2}×2×(1+2)=3$
$(3)x\lt 2$
