$解:(1)∵A(0,3a),B(−4a,0)$
$∴a＞0,OA=3a,OB=4a$
$∴S_{△AOB}=\frac {1}{2}OA×OB=6$
$∴a=1或−1(舍),∴A(0,3),B(−4,0)$
$(2)设A_{1}B_{1}与y轴交于点D,分两种情况:\ $
$当点D在y轴正半轴时,由平移的性质可知$
$A_{1}(m,3−m),B_{1}(−4+m,−m)$
$设直线AB的表达式为y=kx+b,把A(0,3),B(−4,0)代入$
$求得直线AB的表达式为y=\frac{3}{4}+3$
$设线段AB向右平移m个单位长度所得的直线A'B'的$
$表达式为y=\frac{3}{4}x+c$
$与x轴的交点坐标为(−4+m,0),则\frac{3}{4}(−4+m)+c=0$
$解得c=−\frac{3}{4}m+3,∴y=\frac{3}{4}x−\frac{3}{4}m+3$
$∴直线y=\frac {3}{4}x-\frac{3}{4}m+3与y轴的交点为$
$(0,-\frac{3}{4}m+3)$
$∵线段A'B'再向下平移m个单位长度后得到线段A_{1}B_{1}$
$∴D(0,−\frac{3}{4}m+3−m),∴OD=−\frac{3}{4}m+3−m=−\frac{7}{4}m+3$
$∴S_{△A_{1}OB_{1}}=\frac{1}{2}(−\frac{7}{4}m+3)×m+\frac{1}{2}(-\frac {7}{4}m+3)×(4−m)=4$
$解得m=\frac{4}{7}$
$\ ②当点D在y轴负半轴时,由平移的性质可知$
$A_{1}(m,3−m),B_{1}(−4+m,−m),由①得D(0,−\frac{3}{4}m+3−m)$
$∴OD=−(−\frac {3}{4}m+3−m)=\frac{7}{4}m−3$
$∴S_{△A_{1}OB_{1}}=\frac{1}{2}(\frac{7}{4}m−3)×m+\frac{1}{2}(\frac {7}{4}m−3)×(4−m)=4$
$解得m=\frac{20}{7}$
$综上所述,m的值为\frac{4}{7}或\frac{20}{7}$
$(3)由平移的性质,得AB=CD,AB//CD$
$∴∠BAF=∠DCF,∠ABF=∠DCF$
$在△ABF 和△CDF 中$
$\begin{cases}{∠BAF=∠DCF\ } \\ { AB=CD} \\{ ∠ABF=∠CDF} \end{cases}$
$∴ △ABF≌△CDF(ASA),∴AF=CF$
$∴AO−OC=AF+OF−(CF−OF)=AF+OF−CF+OF=2OF$
$∴\frac{AO−OC}{OF}=\frac{2OF}{OF}=2$
$即\frac{AO−OC}{OF}的值为2$
