$解:(1)把A(-6,1)代入y=\frac{m}{x},得1=\frac{m}{-6},∴m=-6$
$∴反比例函数的解析式为y=-\frac{6}{x}$
$把B(1,n)代入得n=-6，∴ B(1,-6)$
$把A(-6,1),B(1,-6)代入y=kx+b得\begin{cases}{ -6k+b=1 }\ \\ { k+b=-6 } \end{cases}解得\begin{cases}{ k=-1 }\ \\ { b=-5 } \end{cases}$
$∴一次函数的解析式为y=-x-5$
$(2)设直线x=-2交直线AB于点H$
$在y=-x-5中令x=-2，得y=-3,∴ H(-2,-3)$
$∵△PAB的面积为 21，∴ \frac{1}{2}PH×|x_{B}-x_{A} |=21,即\frac{1}{2}PH×7=21$
$∴PH=6$
$∵ -3+6=3,-3-6=-9$
$∴点P的坐标为(-2,3)或(-2,-9)$
$(3)(\frac {-11+\sqrt {145}}{2},-\frac {11+\sqrt {145}}{2})或(3,-2)$
