$解:(1)y=2x+6可變形為2x-y+6=0$
$即A=2,B=-1,C=6$
$∴點Q(-1,3)到直線2x-y+6=0的距離$
$d=\frac{|2×(-1)-3+6|}{\sqrt{2^{2}+(-1)^{2}}}=\frac{\sqrt{5}}{5}$
$∴點Q(-1,3)直線y=2x+6的距離為\frac{\sqrt {5}}{5}$
$(2))直線x+\sqrt{6}y=5可變形為y=-\frac {\sqrt {6}}{6}x+\frac {5\sqrt {6}}{6}$
$沿y軸向上平移2個單位長度得到另一條直線y=-\frac {\sqrt {6}}{6}x+\frac {5\sqrt {6}}{6}+2$
$當(dāng)x=0時,y=\frac {5\sqrt {6}}{6}+2,即點(0,\frac{5\sqrt {6}}{6}+2)$
$在直線y=-\frac{\sqrt {6}}{6}x+\frac{5\sqrt {6}}{6}+2上$
$x+ \sqrt{6}=5可變形為x+\sqrt{6}y-5=0$
$∴點(0,\frac{5\sqrt {6}}{6}+2)到直線x+\sqrt {6}y-5=0的距離$
$d=\frac {|1×0+\sqrt {6}×(\frac {5\sqrt {6}}{6}+2)-5|}{\sqrt {1^{2}+(\sqrt {6})^{2}}}=\frac{2\sqrt{42}}{7}$
$∵兩直線平行，∴這兩條直線之間的距離為\frac{2\sqrt{42}}{7}$
