解：?$(1)$?∵?$-2×(-4)+4=12$?
∴集合{?$-4$?，?$12$?}是?$“$?條件集合?$”$?
?$(2)$?∵?$-2×(-\frac {5}{3})+4=\frac {22}{3}$?
∴集合{?$\frac {1}{2}$?，?$-\frac {5}{3}$?，?$\frac {22}{3}$?}是?$“$?條件集合?$”$?
?$(3)$?∵集合{?$8$?，?$n$?}和?$\{m\}$?都是?$“$?條件集合?$”$?，
∴當(dāng)?$-2×8+4=n$?時(shí)，解得?$n=-12$?；當(dāng)?$-2n+4=8$?時(shí)，解得?$n=-2$?；
當(dāng)?$-2n+4=n$?時(shí)，解得?$n=\frac {4}{3}$?；當(dāng)?$-2m+4=m $?時(shí)，解得?$m=\frac {4}{3}$?
綜上，?$m $?的值為?$\frac {4}{3}$?，?$n$?的值為?$-12$?或?$-2$?或?$\frac {4}{3}$?
