18.解:(1)

(2)

当在R上递增,满足题意;

当

∴ ,   ∴

∴ 综上,a的取值范围是.

  19.解法一:

(1) 过O作OF⊥BC于F,连接O1F,

∵OO1⊥面AC,∴BC⊥O1F,

∴∠O1FO是二面角O1-BC-D的平面角,······················· 3分

∵OB = 2,∠OBF = 60°,∴OF =.

在Rt△O1OF中,tan∠O1FO =

∴∠O1FO=60° 即二面角O1-BC-D的大小为60°·············································· 6分

(2) 在△O1AC中,OE是△O1AC的中位线,∴OE∥O1C

∴OE∥O1BC,∵BC⊥面O1OF,∴面O1BC⊥面O1OF,交线O1F.

过O作OH⊥O1F于H,则OH是点O到面O1BC的距离,································· 10分

∴OH = ∴点E到面O1BC的距离等于····················································· 12分

解法二:

(1) ∵OO1⊥平面AC,

∴OO1⊥OA,OO1⊥OB,又OA⊥OB,······················· 2分

建立如图所示的空间直角坐标系(如图)

∵底面ABCD是边长为4,∠DAB = 60°的菱形,

∴OA = 2,OB = 2,

则A(2,0,0),B(0,2,0),C(-2,0,0),O1(0,0,3)·········· 3分

设平面O1BC的法向量为=(x,y,z),则⊥,⊥,

∴,则z = 2,则x=-,y = 3,

∴=(-,3,2),而平面AC的法向量=(0,0,3)··························· 5分

∴ cos<,>=,

设O1-BC-D的平面角为α, ∴cosα=∴α=60°.

故二面角O1-BC-D为60°.············································································· 6分

(2) 设点E到平面O1BC的距离为d,

   ∵E是O1A的中点,∴=(-,0,),············································· 9分

则d=

∴点E到面O1BC的距离等于.··································································· 12分

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