解：?$(1)$?把?$(-1，$??$m)、$??$(2，$??$n)$?代入?$y=x^2+bx-3，$?得?$m=-b-2，$??$n=1+2b$?
∵?$m=n$?
∴?$-b-2=1+2b$?
∴?$b=-1 $?
?$(2)$?∵?$y=x^2-x-3=(x-\frac 12)^2 -\frac {13}{4}$?
∴當(dāng)?$x=\frac {1}{2}$?時(shí)，?$y=-\frac {13}{4}，$?當(dāng)?$x=-3$?時(shí)，?$y=9，$?當(dāng)?$x=2$?時(shí)，?$y=-1$?
∴當(dāng)?$-3< x< 2$?時(shí)，?$y$?的取值范圍為?$- \frac {13}{4} ≤y< 9 $?
?$(3)$?二次函數(shù)?$y=x^2+bx-3$?的對(duì)稱(chēng)軸為直線(xiàn)?$x=- \frac {2}$?
①當(dāng)?$- \frac {2} ≤-1，$?即?$b≥2$?時(shí)，?$x=-1$?的函數(shù)值最小，?$y$?最小?$=-b-2=-5，$?解得?$b=3$?
∴?$y=x^2+3x-3$?
∴當(dāng)?$x=-1$?時(shí)，?$m=-5；$?當(dāng)?$x=2$?時(shí)，?$n=7$?
∴?$m+n=2$?
②當(dāng)?$-1< - \frac {2} < 2，$?即?$-4< b< 2$?時(shí)
?$x=- \frac {2} $?的函數(shù)值最小，?$y$?最小?$=- \frac {1}{4}\ \mathrm ^2-3=-5，$?解得?$b=2 \sqrt{2} ($?舍去)或?$b=-2 \sqrt{2}$?
∴?$y=x^2-2 \sqrt{2} x-3$?
∴當(dāng)?$x=-1$?時(shí)，?$m=2 \sqrt{2} -2；$?當(dāng)?$x=2$?時(shí)，?$n=1-4 \sqrt{2}$?
∴?$m+n=-2 \sqrt{2} -1$?
③當(dāng)?$- \frac {2} ≥2$?即?$b≤-4$?時(shí)，?$x=2$?的函數(shù)值最小，?$y$?最小?$=2b+1=-5，$?解得?$b=-3，$?不滿(mǎn)足?$b≤-4$?
∴此種情況不存在
綜上所述，?$m+n$?的值為?$-2 \sqrt{2} -1$?或?$2$?
