$解:(1)∵A(0,3a),B(?4a,0)$
$∴a＞0,OA=3a,OB=4a$
$∴S_{△AOB}=\frac {1}{2}OA×OB=6$
$∴a=1或?1(舍),∴A(0,3),B(?4,0)$
$(2)設(shè)A_{1}B_{1}與y軸交于點D,分兩種情況:\ $
$當點D在y軸正半軸時,由平移的性質(zhì)可知$
$A_{1}(m,3?m),B_{1}(?4+m,?m)$
$設(shè)直線AB的表達式為y=kx+b,把A(0,3),B(?4,0)代入$
$求得直線AB的表達式為y=\frac{3}{4}+3$
$設(shè)線段AB向右平移m個單位長度所得的直線A'B'的$
$表達式為y=\frac{3}{4}x+c$
$與x軸的交點坐標為(?4+m,0),則\frac{3}{4}(?4+m)+c=0$
$解得c=?\frac{3}{4}m+3,∴y=\frac{3}{4}x?\frac{3}{4}m+3$
$∴直線y=\frac {3}{4}x-\frac{3}{4}m+3與y軸的交點為$
$(0,-\frac{3}{4}m+3)$
$∵線段A'B'再向下平移m個單位長度后得到線段A_{1}B_{1}$
$∴D(0,?\frac{3}{4}m+3?m),∴OD=?\frac{3}{4}m+3?m=?\frac{7}{4}m+3$
$∴S_{△A_{1}OB_{1}}=\frac{1}{2}(?\frac{7}{4}m+3)×m+\frac{1}{2}(-\frac {7}{4}m+3)×(4?m)=4$
$解得m=\frac{4}{7}$
$\ ②當點D在y軸負半軸時,由平移的性質(zhì)可知$
$A_{1}(m,3?m),B_{1}(?4+m,?m),由①得D(0,?\frac{3}{4}m+3?m)$
$∴OD=?(?\frac {3}{4}m+3?m)=\frac{7}{4}m?3$
$∴S_{△A_{1}OB_{1}}=\frac{1}{2}(\frac{7}{4}m?3)×m+\frac{1}{2}(\frac {7}{4}m?3)×(4?m)=4$
$解得m=\frac{20}{7}$
$綜上所述,m的值為\frac{4}{7}或\frac{20}{7}$
$(3)由平移的性質(zhì),得AB=CD,AB//CD$
$∴∠BAF=∠DCF,∠ABF=∠DCF$
$在△ABF 和△CDF 中$
$\begin{cases}{∠BAF=∠DCF\ } \\ { AB=CD} \\{ ∠ABF=∠CDF} \end{cases}$
$∴ △ABF≌△CDF(ASA),∴AF=CF$
$∴AO?OC=AF+OF?(CF?OF)=AF+OF?CF+OF=2OF$
$∴\frac{AO?OC}{OF}=\frac{2OF}{OF}=2$
$即\frac{AO?OC}{OF}的值為2$
