解：?$(1)$?將點(diǎn)?$A(\frac {1}{2}，$??$-\frac {1}{2})$?代入?$y=ax^2$?可得?$-\frac {1}{8}=a×(\frac {1}{2})^2$?
∴?$a=-\frac {1}{2}$?
將點(diǎn)?$B$?代入?$y=-\frac {1}{2}x^2$?可得?$m=-\frac {1}{2}×3^2=-\frac {9}{2}$?
?$(2)$?點(diǎn)?$B$?的坐標(biāo)為?$(3，$??$-\frac {9}{2})，$?函數(shù)關(guān)于?$y$?軸對(duì)稱(chēng)
∴點(diǎn)?$B$?的對(duì)稱(chēng)點(diǎn)的坐標(biāo)為?$(-3，$??$-\frac {9}{2})$?
?$(3)$?∵?$a=-\frac {1}{2}＜0$?
∴當(dāng)?$x＞0$?時(shí)，?$y$?隨?$x$?的增大而減小
?$(4)y=-\frac {1}{2}x^2$?的頂點(diǎn)為?$(0，$??$0)，$?則在?$x=0$?時(shí)，?$y$?取得最大值
