解:\((1)\)如右图.
\((2)\)过\(O\)作\(OH⊥MG\)于点\(H\),设\(DH=xm\),
由\(AB/\!/CD/\!/OH\)得\( \dfrac {MB}{MH}= \dfrac {ND}{NH}\),
即\( \dfrac {1.6}{3.6+x}= \dfrac {0.6}{0.6+x}\),
解得\(x=1.2\).
设\(FG=ym\),同理得\( \dfrac {FG}{HG}= \dfrac {ND}{NH}\),
即\( \dfrac {y}{0.8+y}= \dfrac {0.6}{1.8}\),
解得\(y=0.4\).
所以\(EF\)的影长为\(0.4m\).