解：
(1)显然直线斜率存在.
设在椭圆上的点为(x1,y1)(x2,y2)
那么x1+x2=-4,y1+y2=2
且
x1²/16+y1²/4=1 ①
x2²/16+y2²/4=1 ②
①-②
=> (x1+x2)(x1-x2)/16+(y1+y2)(y1-y2)/4=0
=> -(x1-x2)/4+(y1-y2)/2=0
=> 2(y1-y2)=(x1-x2)
=> (y1-y2)/(x1-x2)=1/2
即斜率为1/2
所以方程为
y-1=(x+2)/2
=> x-2y+4=0

(2)当直线∥y轴时P的轨迹为点(-2,0),当直线∥x轴时P的轨迹为点(0,1)
应该是交于不同的两点A,B
设直线为x=m(y-1)-2,m=(x+2)/(y-1)
A(x1,y1), B(x2,y2), P(x,y)
则x=m(y-1)-2, x²+4y²=16
消去x得(m²+4)y²-(2m²+4m)y+m²+4m-12=0
由于直线与椭圆有两个交点
可得Δ=48m²-64m+192＞0 => 3m²-4m+12＞0该式恒成立.
y=(y1+y2)/2=(2m²+4m)/(2m²+8)=m(m+2)/(m²+4)
x=(x1+x2)/2=m(y1+y2)/2-m-2=-4(m+2)/(m²+4)
y/x=-m/4=-(x+2)/[4(y-1)]
整理得x²+2x+4y²-4y=0
即(x+1)²/2+2(y-1/2)²=1 椭圆
点(-2,0),(0,1)均在椭圆上.
所以点P的轨迹方程为(x+1)²/2+2(y-1/2)²=1