(i) Substitute \(z = 1 + i\) into the equation:

\(w = \frac{1 + i + i}{i(1 + i) + 2} = \frac{1 + 2i}{-1 + i + 2} = \frac{1 + 2i}{1 + i}\).

Multiply numerator and denominator by the conjugate of the denominator:

\(w = \frac{(1 + 2i)(1 - i)}{(1 + i)(1 - i)} = \frac{1 - i + 2i - 2i^2}{1 - i^2} = \frac{1 + i + 2}{2} = \frac{3}{2} + \frac{1}{2}i\).

(ii) Substitute \(w = z\) into the equation:

\(z = \frac{z + i}{iz + 2}\).

Cross-multiply to obtain:

\(z(iz + 2) = z + i\).

\(iz^2 + 2z = z + i\).

\(iz^2 + z - i = 0\).

Substitute \(z = x + iy\) and equate real and imaginary parts:

Real: \(-y^2 + 2x = x\).

Imaginary: \(x^2 + 2y = y - 1\).

Solve these equations to find \(x = -\frac{\sqrt{3}}{2}\) and \(y = \frac{1}{2}\).

Thus, \(z = -\frac{\sqrt{3}}{2} + \frac{1}{2}i\).