Let the included angle θ between OM and X axis, α between OM and tangent, and the slope be k,

Then the included angles between the two tangents and the X axis are θ-α and θ+α, respectively, and let the tangents correspond to slopes k 1, k2,

OM distance is √ (4&; sup2+8 & amp； Sup2)=4√5, the radius of the circle is 2, and the distance from m to the tangent point is 2√ 19.

Then tgα = 2/2 √19 =1√19, and k = tgθ = 8/(-4) =? -2, therefore.

k 1 = TG(θ-α)=(TGθ-TGα)/( 1+TGθTGα)

=(-2- 1/√ 19)/( 1-2/√ 19)=? -(√ 19? +8)/3,

k2 = TG(θ+α)=(TGθ+TGα)/( 1-TGθTGα)

=(-2+ 1/√ 19)/( 1+2/√ 19)=(√ 19? -8)/3,

So the tangent equation is y=? -(√ 19? +8)/3(x+4)+8，

Or y=(√ 19? -8)/3(x+4)+8，

Method 2

Let the tangent be y-8=k(x+4), that is, kx-y+4k+8=0.

The distance from the center of the circle (0,0) to the straight line is | 4k+8 |/√ (1+k&; sup2)=2,

That is 4 & sup2(k+2) and sup2/(1+k&; sup2)-4=0,

4(k & amp； sup2+2k+4)/( 1+k & amp； sup2)- 1=0,

(3k & ampsup2+8k+ 15)/( 1+k & amp； sup2)=0,

There are 3k &；; sup2+8k+ 15=0，

The solution is k = (-8 √ 19)/3.

(2) Let the straight line passing through point n y=k(x-3),

Substitute into the equation of the circle. (In fact, there is no need to substitute for the solution, because in the chord length of the lower region, it takes 4-5k to make sense. sup2>0),

x & ampsup2+k & amp； sup2(x-3)& amp； sup2=4,

That is (1+k&; sup2)x & ampsup2-6k & amp； sup2x+(9k & amp； sup2-4)=0,

To have two intersections, you must △ > 0, that is,

(6k & ampsup2)& ampsup2-4( 1+k & amp； sup2)(9k & ampsup2-4)>0,

The solution is -2/√ 5 < k < 2/√ 5,

The distance from the straight line to the center of the circle is | 3k | ∕√ (1+k&; sup2),

The radius is 2 and the chord length is 2 √ 2&; sup2-(|3k|∕√( 1+k&； sup2))& ampsup2= 2√[(4-5k & amp； sup2)/( 1+k & amp； sup2)],

So s △ OAB = 2 √ [(4-5k&; sup2)/( 1+k & amp； sup2)]*|3k|∕√( 1+k&； sup2)/2 =(3√k & amp； sup2(4-5k & amp； sup2))/( 1+k & amp； sup2)

= 3√4/5-(k & amp； sup2-2/5)& amp； sup2，(-2/√5