解：?$(1)$?∵點(diǎn)?$P$?是線段?$AB$?的黃金分割點(diǎn)，?$AP＜BP$?
∴?$BP=\frac {\sqrt {5}-1}2×AB=\frac {\sqrt 5-1}2=\sqrt 5-1$?
∴?$AP=AB-BP=2-(\sqrt {5}-1)=3-\sqrt {5}$?
?$(2)$?∵?$QP{平分}∠AQB$?
∴?$P$?到?$AQ、$??$BQ$?的距離相等
∴?$\frac {S_{△PAQ}}{S_{△PBQ}}=\frac {AQ}{BQ}=\frac {AP}{PB}$?
又由?$(1)AP=BQ=3-\sqrt {5}$?
∵?$AB=2$?
∴?$PB=AB-AP=2-(3-\sqrt {5})=\sqrt {5}-1$?
∴?$AQ=\frac {AP·BQ}{PB}=\frac {(3-\sqrt 5)^2}{\sqrt 5-1}=2\sqrt {5}-4$?
