(1)：设x=t,y=1则f(t+1)=f(t)+f(1)+3t(t+3)+3
=f(t)+3t(t+3)+4
f(t+1)-f(t)=3t(t+3)+4=3t^2+9t+4
f(2)-f(1)=3*1^2+9*1+4 ,(1)
f(3)-f(2)=3*2^2+9*2+4 ,(2)
f(4)-f(3)=3*3^2+9*3+4 ,(3)
...
f(t)-f(t-1)=3*(t-1)^2+9*(t-1)+4 ,(t-1)
f(t+1)-f(t)=3*t^2+9*t+4 ,(t)
(1)+(2)+(3)+...+(t-1)+(t)
得f(t+1)-f(1)=3*(1^2+2^2+3^2+...t^2)+9*(1+2+3+...+t)+4*t
=t(t+1)(2t+1)/2+9t(t+1)/2+4t
f(t+1)=(t+1-1)(t+1)(2(t+1)-1)/2+9(t+1-1)(t+1)/2+4(t+1-1)+1
所以f(t)=(t-1)t(2t-1)/2+9(t-1)t/2+4(t-1)+1
f(t)=(t-1)t(t+4)+4(t-1)+1
(2)：
(t-1)t(t+4)+4(t-1)+1=t
（t-1)t(t+4)+4(t-1)=t-1
若t1则t^2+4t+3=0
解得t=-1或t=-3
即t为-3,-1,1是首项为-3,公差为2的等差数列
解毕.