(1) 设所求双曲线的标准方程为：x^2/a^2-y^2/b^2=1.
由渐近线方程,得：b/a=±1, b=±a,
且双曲线过(4,-√10), 故4^2/a^2-(-√10)^2/b^2=1.,
16/a^2-10/(a)^2=1.
a^2=6, b^2=a^2=6,
∴x2-y^2=6. ---- (等轴双曲线), 即为所求.
(2).由于c^2=a^2+b^2=6+6=12, c=±2√3.
将M(3,m)代入x^2-y^2=6 中,得：
m=±√3
M(3,√3), 或M(3,-√3).
|MF1|^2=(3+2√3)^2+(√3-0)^2=24+12√3.
|MF2|^2=(3-2√3)^2+(√3-0)^2.=24-12√3..
|F2F1^2|=(2*2√3)^2=48.
|MF1|^2+|MF2|=48.
|MF1|^2+|MF2|^2=|F2F1|^2.
∴MF1⊥MF2.
证毕.