(a) Substitute \(y = 2x + 5\) into \(x^2 + y^2 = 20\) to get \(x^2 + (2x + 5)^2 = 20\).

Expand to get \(x^2 + 4x^2 + 20x + 25 = 20\), leading to \(5x^2 + 20x + 25 = 20\).

Simplify to \(5x^2 + 20x + 5 = 0\) and divide by 5: \(x^2 + 4x + 1 = 0\).

Use the quadratic formula: \(x = \frac{-4 \pm \sqrt{16 - 4}}{2}\).

Calculate \(x = -2 \pm \sqrt{3}\).

Substitute back to find \(y\): \(y = 2(-2 + \sqrt{3}) + 5 = 1 + 2\sqrt{3}\) and \(y = 2(-2 - \sqrt{3}) + 5 = 1 - 2\sqrt{3}\).

Coordinates are \(A = (-2 + \sqrt{3}, 1 + 2\sqrt{3})\) and \(B = (-2 - \sqrt{3}, 1 - 2\sqrt{3})\).

Use the distance formula: \(AB = \sqrt{((-2 + \sqrt{3}) - (-2 - \sqrt{3}))^2 + ((1 + 2\sqrt{3}) - (1 - 2\sqrt{3}))^2}\).

Simplify to \(AB = \sqrt{(2\sqrt{3})^2 + (4\sqrt{3})^2} = \sqrt{12 + 48} = \sqrt{60}\).

(b) The equation of the tangent is \(y = m(x - 10)\).

Substitute into the circle equation: \(x^2 + (m(x - 10))^2 = 20\).

Expand to get \(x^2 + m^2(x^2 - 20x + 100) = 20\).

Collect terms: \((m^2 + 1)x^2 - 20m^2x + 100m^2 = 20\).

Use the discriminant condition for tangency: \(b^2 - 4ac = 0\).

Calculate \(400m^4 - 80(5m^2 + 4m^2 - 1) = 0\).

Simplify to \(-80(4m^2 - 1) = 0\), leading to \(m = \pm \frac{1}{2}\).