(i) To find the value of \(b\), we need to find the integer root of the polynomial \(x^4 - 2x^3 - 7x - 6 = 0\). By inspection or using the Rational Root Theorem, we find that \(b = 3\) is a root.

(ii) To show that \(a\) satisfies the equation \(a = -\frac{1}{3}(2 + a^2 + a^3)\), perform polynomial division of \(x^4 - 2x^3 - 7x - 6\) by \(x - 3\). The quotient is \(x^3 + x^2 + 3x + 2\). Setting this equal to zero gives \(x^3 + x^2 + 3x + 2 = 0\). Rearranging gives \(a = -\frac{1}{3}(2 + a^2 + a^3)\).

(iii) Using the iterative formula \(a_{n+1} = -\frac{1}{3}(2 + a_n^2 + a_n^3)\), start with an initial guess, say \(a_0 = -1\). Calculate successive iterations:

\(a_1 = -\frac{1}{3}(2 + (-1)^2 + (-1)^3) = -0.66667\)

\(a_2 = -\frac{1}{3}(2 + (-0.66667)^2 + (-0.66667)^3) = -0.71605\)

\(a_3 = -\frac{1}{3}(2 + (-0.71605)^2 + (-0.71605)^3) = -0.71484\)

\(a_4 = -\frac{1}{3}(2 + (-0.71484)^2 + (-0.71484)^3) = -0.71506\)

\(a_5 = -\frac{1}{3}(2 + (-0.71506)^2 + (-0.71506)^3) = -0.71498\)

Continue until the value stabilizes to \(a \approx -0.715\) to 3 decimal places.