解:?$(1)2(mx^2-x-\frac 72)+4x^2+3nx$?
?$=2\ \mathrm {m}x^2-2x-7+4x^2+3nx$?
?$=(2\ \mathrm {m}+4)x^2+(3n-2)x-7$?
∴?$2\ \mathrm {m}+4=0,$??$3n-2=0$?
∴?$m=-2,$??$n=\frac 23$?
?$(2)$?原式?$=6\ \mathrm {m^2}-9\ \mathrm {m}n-15\ \mathrm {m}-3-6\ \mathrm {m^2}+6\ \mathrm {m}n-6$?
?$=(6-6)\mathrm {m^2}+(-9+6)mn-15\ \mathrm {m}+(-3-6)$?
?$=-3\ \mathrm {m}n-15\ \mathrm {m}-9$?
?$=-3×(-2)×\frac 23-15×(-2)-9$?
?$=4+30-9$?
?$=25$?