(a) To show that the equation \(x^5 = 2 + x\) has exactly one real root, sketch the graphs of \(y = x^5\) and \(y = x + 2\). The intersection of these graphs represents the solution to the equation. Since \(x^5\) is a strictly increasing function and \(x + 2\) is a linear function, they intersect at exactly one point, confirming one real root.

(b) Consider the iterative formula \(x_{n+1} = \frac{4x_n^5 + 2}{5x_n^4 - 1}\). If this sequence converges, it must satisfy the equation \(x = \frac{4x^5 + 2}{5x^4 - 1}\). Rearranging gives \(x^5 = 2 + x\), which is the equation from part (a). Therefore, if the sequence converges, it converges to the root of the equation.

(c) Using the iterative formula with \(x_1 = 1.5\):

1. Calculate \(x_2 = \frac{4(1.5)^5 + 2}{5(1.5)^4 - 1} = 1.27103\).

2. Calculate \(x_3 = \frac{4(1.27103)^5 + 2}{5(1.27103)^4 - 1} = 1.26713\).

3. Calculate \(x_4 = \frac{4(1.26713)^5 + 2}{5(1.26713)^4 - 1} = 1.26700\).

4. Calculate \(x_5 = \frac{4(1.26700)^5 + 2}{5(1.26700)^4 - 1} = 1.26700\).

The sequence converges to 1.267, correct to 3 decimal places.