已知点An(an,bn)在曲线y=e的x次方上,a1=a,Sn是各项都不为零的...

：(1)证明：b(n+2)/bn=e^a(n+2)/e^an=e^[a(n+2)-an]
要证明{b(n+2)/bn}为常数数列
,只需证a(n+2)-an为常数；
∵Sn^2=3n^2*an+S(n-1)^2
∴Sn^2-S(n-1)^2=[Sn+S(n-1)][Sn-S(n-1)]=[Sn+S(n-1)]*an=3n^2*an
∴Sn+S(n-1)=3n^2……①,S(n+1)+Sn=3(n+1)^2……②,S(n+2)+S(n+1)=3(n+2)^2……③
②-①：S(n+1)-S(n-1)=6n+3,即a(n+1)+an=6n+3……④
③-②：S(n+2)-Sn=6n+9,即a(n+2)+a(n+1)=6n+9……⑤
⑤-④：a(n+2)-an=6n+9-6n-3=6……⑥
∴a(n+2)-an=6为常数,数列{b(n+2)/bn}为常数列,且b(n+2)/bn=e^6
(2)由①：S2+S1=12,a2+2a1=12
∴a2=12-2a1=12-2a；由④：a3+a2=15,a4+a3=21
∴a3=15-a2=3+2a；a4=21-a3=18-2a
由⑥可看出：数列{a2k}、{a(2k+1)}(k属于Z+)分别是以a2、a3为首项,6为公差的等差数列
∴a2k=a2+6(k-1),a(2k+1)=a3+6(k-1),a(2k+2)=a4+6(k-1)(k属于N*)
数列{an}为单调递增数列a1＜a2且a2k＜a(2k+1)＜a(2k+2)对任意k属于N*成立
a1＜a2且a2+6(k-1)＜a3+6(k-1)＜a4+6(k-1)a1＜a2＜a3＜a4
a1＜12-2a＜3+2a＜18-2a9/4＜a＜15/4
∴M={a|9/4＜a＜15/4}