解：?$(2)$?∵?$(k$?，?$-3)$?是?$“$?積差等數(shù)對(duì)?$”$?
∴?$k-(-3)=-3k$?，解得：?$k=-\frac 34$?
?$(3)$?原式?$=4(3mn-m-2mn+2)-6\ \mathrm {m^2}+4n+6\ \mathrm {m^2}$?
?$=12mn-4m-8mn+8-6\ \mathrm {m^2}+4n+6\ \mathrm {m^2}$?
?$=4mn-4m+4n+8$?
∵?$(m$?，?$n)$?是?$“$?積差等數(shù)對(duì)?$”$?
∴?$m-n=mn$?
∴原式?$=4mn-4(m-n)+8=4mn-4mn+8=8$?
