设M(-1/2,m),A(-1/2-h,m-k),B(-1/2+h,m+k),
A,B在椭圆x^2/2+y^2=1上,
∴(-1/2-h)^2/2+(m-k)^2=1,①
（-1/2+h)^2/2+(m+k)^2=1,②
①-②,h-4mk=0,
∴AB的斜率=k/h=1/(4m),
PQ⊥AB,∴PQ的斜率=-4m,
∴PQ:y-m=-4m(x+1/2),即y=-4mx-m,③
代入椭圆方程得x^2+2(16m^2x^2+8m^2x+m^2)=2,
整理得(1+32m^2)x^2+16m^2x+2m^2-2=0,
F2（1,0）,设P(x1,y1),Q(x2,y2),则x1+x2=-16m^2/(1+32m^2),x1x2=(2m^2-2)/(1+32m^2),
由③,y1y2=(-4mx1-m)(-4mx2-m)=m^2[16x1x2+4(x1+x2)+1],
∴向量F2P*F2Q=(x1-1,y1)*(x2-1,y2)=(x1-1)(x2-1)+y1y2
=x1x2-(x1+x2)+1+y1y2
=(1+16m^2)x1x2+(4m^2-1)(x1+x2)+1+m^2
=[(1+16m^2)(2m^2-2)-16m^2(4m^2-1)+(1+m^2)(1+32m^2)]/(1+32m^2)
=[32m^4-30m^2-2
-64m^4+16m^2
+32m^4+33m^2+1]/(1+32m^2)
=(19m^2-1)/(32m^2+1),
M在椭圆内,∴1/8+m^2