(a) \(C=(8,4)\), radius 10. \(CT^2=(-6-8)^2+(6-4)^2=200\), so \(CT=10\sqrt2>10\). Therefore \(T\) is outside.

(b) Radius \(CA\) is perpendicular to tangent \(TA\). In the right triangle \(CTA\), \[\sin\angle CTA=\frac{CA}{CT}=\frac{10}{10\sqrt2}=\frac1{\sqrt2},\] so \(\angle CTA=45^\circ\).

(c) The right triangle has equal legs \(CA=TA=10\). Its perpendicular from \(A\) to \(CT\) bisects the hypotenuse; by symmetry \(AB\) passes through this midpoint \(M=(1,5)\), perpendicular to \(CT\). The gradient of \(CT\) is \(-1/7\), hence that of \(AB\) is 7: \(y-5=7(x-1)\), giving \(\boxed{y=7x-2}\).

(d) Substitute into the circle: \[(x-8)^2+(7x-6)^2=100\quad\Longrightarrow\quad50x^2-100x=0.\] Thus the required coordinates are \(\boxed{x=0\text{ and }x=2}\).