（1）b=1,a^2=2,椭圆方程是x^2/2+y^2=1.①
(2)设l:y=kx-1/3,②代入①,
x^2/2+k^2x^2-(2/3)kx+1/9=1,
(k^2+1/2)x^2-(2/3)kx-8/9=0,
设交点A(x1,y1),B(x2,y2),则x1+x2=(2k/3)/(k^2+1/2),x1x2=(-8/9)/(k^2+1/2),
由②,y1+y2=k(x1+x2)-2/3,
y1y2=(kx1-1/3)(kx2-1/3)=k^2x1x2-(k/3)(x1+x2)+1/9,
设T(s,t),则向量AT×BT=(s-x1)(s-x2)+(t-y1)(t-y2)
=s^2-s(x1+x2)+x1x2+t^2-t(y1+y2)+y1y2
=s^2+t^2+2t/3+1/9-(s+k/3+kt)(x1+x2)+(1+k^2)x1x2=0,
∴(s^2+t^2+2t/3+1/9)(k^2+1/2)-(s+k/3+kt)(2k/3)-(8/9)(1+k^2)=0,
整理得(s^2+t^2+2t/3+1/9-2/9-2t/3-8/9)k^2-(2s/3)k+(s^2+t^2+2t/3+1/9)/2-8/9=0,对k恒成立,
∴s^2+t^2-1=0,s=0,(s^2+t^2+2t/3+1/9)/2-8/9=0,
第一、三式变为t^2=1,t^2+2t/3-5/3=0,解得t=1.
∴存在定点T(0,1).