According to the length of two tangents drawn at a point outside the circle and the definition of hyperbola, we can get.

|PF 1|-|PF2|=2a。

Because | nf1| | nf2 | = | pf1| | pf2 | = 2a. ①

|NF 1|+|NF2|=2c。 ②

| nf 1 | = = a+c from ① ②.

∴|on|=|nf 1|-|of 1|=a+c-c=a=3.

Therefore, the coordinate of the tangent point n is (3,0).

According to symmetry, when p is on the left branch of hyperbola, the coordinate of the tangent point n is (-3,0).

(2)a=3，b=2

So c=√ 13.

Might as well set pf1> PF2

Let pf 1 = m and pf2 = n.

It is defined by hyperbola.

m-n=2a=6

mn=32

So (m-n) 2 = m 2+n 2-2mn = 36.

m^2+n^2= 100

F 1F2=2c=2√ 13

So in the triangle PF 1F2

Cos angle f1pf 2 = (m2+N2-f1F2)/2mn.

=( 100-52)/64

=3/4

So the angle f1pf 2 = arccos 3/4 = 41.41degree.