(a) Substitute \(w = x + iy\) and \(w^* = x - iy\) into the equation:

\((1 + 2i)(x + iy) + i(x - iy) = 3 + 5i\)

Expand and simplify:

\((1 + 2i)(x + iy) = x + 2ix + iy - 2y = (x - 2y) + i(2x + y)\)

\(i(x - iy) = ix + y\)

Combine terms:

\((x - 2y) + i(2x + y) + ix + y = 3 + 5i\)

Equate real and imaginary parts:

Real: \(x - 2y + y = 3\)

Imaginary: \(2x + y + x = 5\)

Simplify to get two equations:

\(x - y = 3\)

\(3x + y = 5\)

Solve these equations to find \(x = 2\) and \(y = -1\).

Thus, \(z = 2 - i\).

(b)(ii) The region is a circle centered at \(2 + 2i\) with radius 1, intersecting the line \(\arg(z - 4i) = -\frac{1}{4}\pi\).

The least value of \(\text{Im } z\) is at the lowest point of the circle, which is \(2 - \frac{1}{2}\sqrt{2}\).