(a) To find \(\frac{dy}{dx}\), use the product rule. Let \(u = x\) and \(v = \arctan\left(\frac{1}{2}x\right)\).

Then \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = \frac{1}{1 + \left(\frac{1}{2}x\right)^2} \cdot \frac{1}{2} = \frac{1}{4 + x^2}\).

Using the product rule: \(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = x \cdot \frac{1}{4 + x^2} + \arctan\left(\frac{1}{2}x\right)\).

Thus, \(\frac{dy}{dx} = \arctan\left(\frac{1}{2}x\right) + \frac{2x}{x^2 + 4}\).

(b) At \(x = 2\), the y-coordinate is \(y = 2 \arctan(1) = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}\).

The gradient at \(x = 2\) is \(\frac{dy}{dx} = \arctan(1) + \frac{4}{8} = \frac{\pi}{4} + \frac{1}{2}\).

The equation of the tangent is \(y - \frac{\pi}{2} = \left(\frac{\pi}{4} + \frac{1}{2}\right)(x - 2)\).

Setting \(x = 0\) to find \(p\):

\(p - \frac{\pi}{2} = \left(\frac{\pi}{4} + \frac{1}{2}\right)(-2)\)

\(p = \frac{\pi}{2} - \left(\frac{\pi}{2} + 1\right) = -1\).