(a) To find \(\frac{dy}{dx}\), we use the chain rule:

\(\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}\).

First, find \(\frac{dx}{dt}\):

\(x = \sqrt{t} + 3 \Rightarrow \frac{dx}{dt} = \frac{1}{2\sqrt{t}}\).

Next, find \(\frac{dy}{dt}\):

\(y = \ln t \Rightarrow \frac{dy}{dt} = \frac{1}{t}\).

Thus, \(\frac{dy}{dx} = \frac{1}{t} \times 2\sqrt{t} = \frac{2}{\sqrt{t}}\).

(b) The gradient of the normal is \(-2\), so the gradient of the tangent is \(\frac{1}{2}\).

Set \(\frac{dy}{dx} = \frac{1}{2}\):

\(\frac{2}{\sqrt{t}} = \frac{1}{2} \Rightarrow \sqrt{t} = 4 \Rightarrow t = 16\).

Substitute \(t = 16\) into the parametric equations:

\(x = \sqrt{16} + 3 = 7\),

\(y = \ln 16\).

Therefore, the coordinates are \((7, \ln 16)\).