解：??$(1) $??設(shè)??$ y=k(x-3)$??，
把??$ x=4$??，??$y=3 $??代入得??$ k=3$??，
∴??$y $??與??$ x $??的函數(shù)表達(dá)式為??$ y=3(x-3)$??，即??$ y=3 x-9$??
??$(2) $??設(shè)??$ y-1=k x$??，
把??$ x=2$??，??$y=-4 $??代入得??$ -4-1=2\ \mathrm {k}$??，解得??$ k=-\frac {5}{2}$??
∴??$y-1=-\frac {5}{2} x$??
∴??$y $??與??$ x $??的函數(shù)表達(dá)式為??$ y=-\frac {5}{2} x+1$??；
??$(3) $??設(shè)??$ y-{1}=a x$??，??$y-{2}=b(x-2)$??，則??$ y=a x+b(x-2)$??，
把??$ x=-1$??，??$y=2$??和??$x=2$??，??$y=5 $??代入得
??$\begin {cases}{-a-3\ \mathrm =2}\\{2a=5}\end {cases}$??，解得??$\begin {cases}{a=\frac {5}{2}}\\{b=-\frac {3}{2}}\end {cases}$??
∴??$y=\frac {5}{2} x-\frac {3}{2}(x-2)=x+3$??，
∴??$y $??與??$ x $??的函數(shù)表達(dá)式為??$ y=x+3.$??
