$解:(3)如图②,线段DF即为小亮站在D处的影子$
$设OP=x m,由题意,易得AB//PO,∴△EAB∽△EPO$
$∴\frac{AB}{PO}=\frac{BE}{OE},即\frac{1.6}{x}=\frac{1.6}{4.2+1.6},解得x=5.8$
$∴OP=5.8m,当OD=6m时,设小亮的影长DF为y m$
$则OF=OD+DF=(6+y)m$
$由题意,易得CD//PO,∴△FCD∽△FPO$
$∴\frac{DF}{OF}=\frac{CD}{PO},即\frac{y}{6+y}=\frac{1.6}{5.8},解得=\frac{16}{7}$
$∴DF=\frac{16}{7}m$
$∴小亮的影长是\frac{16}{7} m$