（1）双曲线X^2-Y^2/3=1的焦点是（土2,0),
设所求椭圆方程是x^2/(b^2+4)+y^2/b^2=1,
它过点(2,3),
∴4/(b^2+4)+9/b^2=1,
4b^2+9b^2+36=b^4+4b^2,
b^4-9b^2-36=0,b^2>0,
∴b^2=12,
∴椭圆方程是x^2/16+y^2/12=1.其左右顶点分别是A(-4,0),B(4,0),右焦点F(2,0),N(8,n),
AN:y=n(x+4)/12与椭圆交于M(4(108-n^2)/(108+n^2),72n/(108+n^2)),
AM=MN,
[4(108-n^2)/(108+n^2)+4]^2+[72n/(108+n^2)]^2
=[4(108-n^2)/(108+n^2)-8]^2+[72n/(108+n^2)-n]^2,
∴12[8(108-n^2)/(108+n^2)-4]+n[144n/(108+n^2)-n]=0,
48(108-3n^2)+n^2(36-n^2)=0,
n^4+108n^2-48*108=0,
∴n^2=36,n=土6.M(2,土3）,
当M(2,3)时MA=(-6,-3),MB=(2,-3),|MA|=3√5,|MB|=√13,MA*MB=-3,
cosAMB=-√65/65,
当M(2,-3)时cosAMB=-√65/65.
（2）?