解:?$(1)$?以?$C$?為圓心,?$CM$?長(zhǎng)為半徑畫圓,
連接?$CN$?交?$DE$?于?$M_1,$?延長(zhǎng)?$NC$?交圓于?$M_2$?
∵?$△ACB$?是等腰直角三角形,?$N$?是?$AB$?中點(diǎn)
∴?$CN$?平分?$∠ACB,$??$CN=\frac {1}{2}AB=\frac {1}{2}×4=2$?
∵?$△DCE$?是等腰直角三角形,?$M_1$?是?$DE$?中點(diǎn)
∴?$CM_1=\frac {1}{2}DE=\frac {1}{2}×2=1$?
∴?$M、$??$N$?距離的最小值是?$NM_1=CN-CM_1=2-1=1$?
?$M、$??$N$?距離的最大值是?$NM_2=CN+CM_2=2+1=3$?
?$(2)$?連接?$CM,$??$CN,$?作?$NH⊥MC$?交?$MC$?延長(zhǎng)線于?$H$?
∵?$△ACB$?是等腰直角三角形,?$N$?是?$AB$?中點(diǎn)
∴?$CN=\frac {1}{2}AB=2$?
同理:?$CM=\frac {1}{2}DE=1$?
∵?$△CDE$?繞頂點(diǎn)?$C$?逆時(shí)針旋轉(zhuǎn)?$120°$?
∴?$∠MCN=120°$?
∴?$∠NCH=180°-∠MCN=60°$?
∴?$CH=\frac {1}{2}CN=1$?
∴?$NH=\sqrt {3}CH=\sqrt {3}$?
∵?$MH=MC+CH=2$?
∴?$MN=\sqrt {MH^2+NH^2}=\sqrt {7}$?