(1)解由题意可知,1-a_(n+1)^2=2/3(1-a_n^2),

　　令cn=1-a_n^2,则cn+1=2/3 cn,

　　又c1=1-a_1^2=3/4,

　　则数列{cn}是首项为c1=3/4,公比为 2/3 的等比数列,

　　即cn=3/4×(2/3)^(n"-" 1),故1-a_n^2=3/4·(2/3)^(n"-" 1),

　　∴a_n^2=1-3/4·(2/3)^(n"-" 1).

　　又a1=1/2>0,anan+1<0,

　　故an=(-1)n-1√(1"-" 3/4 "·" (2/3)^(n"-" 1) )(n∈N+),

　　bn=a_(n+1)^2-a_n^2

　　=[1"-" 3/4 "·" (2/3)^n ]-[1"-" 3/4 "·" (2/3)^(n"-" 1) ]

　　=1/4·(2/3)^(n"-" 1) (n∈N+).

(2)证明用反证法证明.

　　假设数列{bn}中存在三项br,bs,bt(rbs>bt,则只可能有2bs=br+bt成立.

　　即2×1/4·(2/3)^(s"-" 1)=〖1/4 "·" (2/3)〗^(r"-" 1)+1/4·(2/3)^(t"-" 1),两边同乘3t-121-r化简,得3t-r+2t-r=2·2s-r3t-s,因为r