(理)解: (1)設0≤x1<x2≤1,則必存在實數tÎ(0,1),使得x2=x1+t,
由條件③得,f(x2)=f(x1+t)³f(x1)+f(t)-2,
∴f(x2)-f(x1)³f(t)-2,
由條件②得, f(x2)-f(x1)³0,
故當0≤x≤1時,有f(0)≤f(x)≤f(1).
又在條件③中,令x1=0,x2=1,得f(1)³f(1)+f(0)-2,即f(0)≤2,∴f(0)=2,
故函數f(x)的最大值為3,最小值為2.
(2)解:在條件③中,令x1=x2=,得f()³2f()-2,即f()-2≤[f()-2],
故當nÎN*時,有f()-2≤[f()-2]≤[f()-2]≤···≤[f()-2]=,
即f()≤+2.
又f()=f(1)=3≤2+,所以對一切nÎN,都有f()≤+2.
(3)對一切xÎ(0,1

,都有

.對任意滿足xÎ(0,1

,總存在n(nÎN),使得
<x≤, 根據(1)(2)結論,可知:f(x)≤f()≤+2,
且2x+2>2´+2=+2,故有

.
綜上所述,對任意xÎ(0,1

,

恒成立.