解：$(1)(5\ \mathrm {m^2}-4m+2)-(4\ \mathrm {m^2}-4m-7)=5\ \mathrm {m^2}-4m+2-4\ \mathrm {m^2}+4m+7=\mathrm {m^2}+9$
∵不论$m $为何值，$\mathrm {m^2}+9＞0$恒成立
∴$5\ \mathrm {m^2}-4m+2＞4\ \mathrm {m^2}-4m-7$
$(2)$∵$A=5\ \mathrm {m^2}-4(\frac {7}{4}m -\frac {1}{2})$，$B=7(\mathrm {m^2}-m)+3$
∴$A-B=[5\ \mathrm {m^2}-4(\frac {7}{4}m -\frac {1}{2}) -[7(m²-m)+3]$
$=5\ \mathrm {m^2}-4(\frac {7}{4}m -\frac {1}{2})-7(\mathrm {m^2}-m)-3$
$=5\ \mathrm {m^2}-7m+2-7\ \mathrm {m^2}+7m-3$
$=-2\ \mathrm {m^2}-1$
∵不论$m $为何值，$-2\ \mathrm {m^2}-1＜0$恒成立
∴$A-B＜0$，即$A＜B$
$(3)(3a+2b)-(2a+3b)=3a+2b-2a-3b=a-b$
当$a＞b$时，$a-b＞0$，此时$3a+2b＞2a+3b$
$ $当$a=b$时，$a-b=0$，此时$3a+2b=2a+3b$
当$a＜b$时，$a-b＜0$，此时$3a+2b＜2a+3b$
