Solution: (1) circle c: x 2+y 2+2mx-6my+ 1 = 0,? (x+m)^2+(y-3m)^2= 10m^2- 1，

The coordinate of the center point c is (-m, 3m). The square of the circle radius C = 10m 2- 1,

If two points P and Q on the circle C are symmetrical about the straight line x-y+4=0, then the center C(-m, 3m) must be on the straight line x-y+4=0.

So: -m-3m+4=0, m= 1.

? (2) ? m= 1，？ The coordinates of point C are (-1, 3). The radius of a circle C = 3 under the radical sign (10m2-1).

As shown in the figure:

? Points P and Q are symmetrical about the straight line y=x+4, and the straight line PQ is perpendicular to the straight line y=x+4, and the circle with PQ as its diameter (tentatively designated as circle A).

? The center of circle A is the intersection of line PQ and line y=x+4, so that the coordinate of center A is (x, x+4), circle A passes through the origin O, and the line segment?

? AO and AP are the radii of circle a, AO=AP, ao 2 = AP 2,?

In circle c, CP is the radius, CP=3, AP 2 = CP 2-CA 2,? So what? CP^2-CA^2=AO^2，

3^2-[(- 1-x)^2+(3-x-4)^2]=x^2+(x+4)^2？ X=-3/2，x+4=5/2，？ The coordinate of point A are (-3/2, 5/2),

The straight line PQ is perpendicular to the straight line y=x+4, and the slope of the straight line PQ is-1, so the analytical formula of the straight line PQ is y=-x+b, and the point A is in the straight line PQ.

Up，5/2=-(-3/2)+b，？ b= 1，

So the analytical formula of direct PQ is? y=-x+ 1，？ That is x+y- 1=0.

(Using the vertical diameter theorem and the distance formula between two points, the method is simple and easy to understand, and the calculation amount is small)

Hope to adopt!