x 1^2-y 1^2/2= 1

x2^2-y2^2/2= 1

The two expressions are subtracted to get (x1+x2) (x1-x2)-(y1+y2) (y1-y2)/2 = 0.

If P( 1, 2) is the midpoint of AB, then x 1+x2=2, y 1+y2=4.

Therefore, k (ab) = (y1-y2)/(x1-x2) = 2 (x1+x2)/(y1+y2) =1.

So the equation of AB is y-2= 1*(x- 1), that is, y=x+ 1.

(2) Suppose there is a chord with q as the midpoint, and the coordinates of chord AB are (x 1, y 1) and (x2, y2) respectively.

As above, there is k (ab) = 2 (x1+x2)/(y1+y2) = 2 * 2/2 = 2.

So the AB equation is y- 1=2(x- 1), that is, y=2x- 1.

Substituting hyperbola is x 2-(2x- 1) 2/2 = 1.

2x^2-(4x^2-4x+ 1)=2

2x^2-4x+3=0

Discriminant = 4 2-4 * 2 * 3

So the equation has no solution, that is to say, the string with q as the midpoint does not exist.