(a) Substitute the line into the circle: \[x^2+(2x+5)^2=20\quad\Longrightarrow\quad x^2+4x+1=0.\] Thus \(x=-2\pm\sqrt3\) and \(y=1\pm2\sqrt3\), with corresponding signs. Hence \[\boxed{A=(-2+\sqrt3,1+2\sqrt3),\quad B=(-2-\sqrt3,1-2\sqrt3)}.\] The differences of coordinates are \(2\sqrt3\) and \(4\sqrt3\), so \[\boxed{AB=\sqrt{(2\sqrt3)^2+(4\sqrt3)^2}=\sqrt{60}=2\sqrt{15}}.\]

(b) The tangent has equation \(y=m(x-10)\). Substitution into the circle gives \[(1+m^2)x^2-20m^2x+100m^2-20=0.\] Tangency requires a repeated root, so its discriminant is zero: \[400m^4-4(1+m^2)(100m^2-20)=80(1-4m^2)=0.\] Thus \[\boxed{m=\frac12\quad\text{or}\quad m=-\frac12}.\]