The second question:

( 1)

It can be seen from the question that (a+√3b)(a-√3b)=0.

Namely: (x+√3, √3y)(x-√3, -√3y)=0.

De: x2-3-3y2=0

So the trajectory equation of point P(x, y) is

x2/3-y2= 1

(2) 1. When the straight line L intersects with D(0,-1)

- 1=k*0+m

At this time m=- 1.

2. When the straight line L intersects the hyperbola.

By x2-3-3y2=0

Y=kx+m, you can get (1/3-k2) x2-2kmx-m2-1= 0 by eliminating y.

Let A(x, y), B(X, y), AB midpoint c.

xX =(-3 m2-3)/( 1-3 k2)÷0……①

△= 4k2m 2+4(m2+ 1)( 1/3-k2)÷0……②

X+x = 6km /( 1-3k2)

So the midpoint C{3km/( 1-3k2), m/( 1-3k2)}

So the perpendicular equation of AB is y-m/(1-3k2) =-1/k {x-3km/(1-3k2)}.

Because it passed D(0,-1)

Then:-1-m/(1-3k2) = 3km/(1-3k2), that is: 4m/(1-3k2) =-1... ③.

It can be obtained from ① ② ③: m > 4.

To sum up, the value range of m is {- 1} ∨( 4, +∞).