(1)∵S[n]是数列{a[n]}(n∈N+）的前n项和
且S[n]^2=3n^2*a[n]+S[n-1]^2,a[n]≠0,n=2,3,4,...
∴S[n]^2-S[n-1]^2=3n^2*a[n]
(S[n]+S[n-1])(S[n]-S[n-1])=3n^2*a[n]
(S[n]+S[n-1])a[n]=3n^2*a[n]
S[n]+S[n-1]=3n^2
∵a[1]=1
S[2]+S[1]=a[2]+2a[1]=3*2^2
∴a[2]=10
∵S[3]+S[2]=a[3]+2(a[2]+a[1])=3*3^2
∴a[3]=5
(2)∵S[n]+S[n-1]=3n^2
∴S[n+1]+S[n]=3(n+1)^2
将上面两式相减,得：
a[n+1]+a[n]=6n+3 (n=2,3,4,...)
用待定系数法,得：
a[n+1]+x(n+1)+y=-(a[n]+xn+y)
∵-2x=6,-x-2y=3
∴x=-3,y=0
∴a[n+1]-3(n+1)=-(a[n]-3n) (n=2,3,4,...)
∵a[2]=10
∴{a[n]-3n}是首项为a[2]-3*2=4,公比为-1的等比数列
即：a[n]-3n=4(-1)^(n-2)=4(-1)^n
∴a[n]=4(-1)^n+3n (n=2,3,4,...)
∵n=1代入上式,得：a[1]=7≠1,即a[1]不符合上式
∴{a[n]}的通项公式是：
a[n]=1 (n=1)
a[n]=4(-1)^n+3n (n=2,3,4,...)