解：?$(2)$?由題意可得?$4=x^2-3x$?
∴?$\frac 23x^2-2x+3=\frac 23(x^2-3x)+3$?
?$=\frac 23×4+3=\frac {17}{3}$?
?$(3)$?由題意可得?$p=\mathrm {m^2}-n$?，?$q=n^2-p$?
∴?$q=n^2-(\mathrm {m^2}-n)=n^2-\mathrm {m^2}+n$?
∵?$p-\frac 12q=2$?
∴?$2p-q=4$?
∴?$2(\mathrm {m^2}-n)-(n^2-\mathrm {m^2}+n)=4$?
∴?$2\ \mathrm {m^2}-2n-n^2+\mathrm {m^2}-n=4$?
∴?$3\ \mathrm {m^2}-n^2-3n=4$?
∴?$n^2-3\ \mathrm {m^2}+4=-3n$?，?$n^2-3\ \mathrm {m^2}+3n=-4$?
∴原式?$=n · (-3n)+9(\mathrm {m^2}-n)+2012$?
?$=-3n^2+9\ \mathrm {m^2}-9n+2012$?
?$=-3(n^2-3\ \mathrm {m^2}+3n)+2012$?
?$=-3×(-4)+2012$?
?$=12+2012$?
?$=2024$?
