(a) Sketch the graphs of \(y = e^x - 3\) and \(y = \sqrt{x}\). The graph of \(y = e^x - 3\) cuts the vertical axis at (0, -2) and has an increasing gradient. The graph of \(y = \sqrt{x}\) starts at (0, 0) and has a reducing gradient. The intersection of these graphs indicates the root.

(b) Calculate \(e^1 - 3 = -0.28\) and \(e^2 - 3 = 4.39\). Since \(\sqrt{1} = 1\) and \(\sqrt{2} \approx 1.41\), we have \(1 > -0.28\) and \(1.41 < 4.39\). Therefore, the root lies between 1 and 2.

(c) Rearrange \(\sqrt{x} = e^x - 3\) to \(x = \ln(3 + \sqrt{x})\). The iterative formula \(x_{n+1} = \ln(3 + \sqrt{x_n})\) converges to the root of the equation.

(d) Using the iterative formula, start with an initial guess, e.g., \(x_1 = 1\):

\(x_2 = \ln(3 + \sqrt{1}) = 1.3864\)

\(x_3 = \ln(3 + \sqrt{1.3864}) = 1.4297\)

\(x_4 = \ln(3 + \sqrt{1.4297}) = 1.4341\)

\(x_5 = \ln(3 + \sqrt{1.4341}) = 1.4344\)

\(x_6 = \ln(3 + \sqrt{1.4344}) = 1.4345\)

The root correct to 2 decimal places is 1.43.