Do not use a calculator in answering this question. The complex numbers $z$ and $w$ satisfy the following equations. $w+z^{*}=-2+4i$ $z+2=3iw$ Find $z$ and $w$, giving your answers in the form $a+ib$, where $a$ and $b$ are real numbers.

1. Express $w$ from the first equation and substitute it into the second equation.

From the first equation, express $w$:
$w = -2 + 4i - z^*$

Substitute this expression for $w$ into the second equation:
$z + 2 = 3i(-2 + 4i - z^*)$
$z + 2 = -6i + 12i^2 - 3iz^*$
$z + 3iz^* = -14 - 6i$

2. Substitute the Cartesian form $z = a + ib$ to separate the real and imaginary parts.

Let $z = a + ib$, then the conjugate is $z^* = a - ib$.
Substitute into the equation:
$(a + ib) + 3i(a - ib) = -14 - 6i$
$a + ib + 3ai - 3i^2 b = -14 - 6i$
$a + ib + 3ai + 3b = -14 - 6i$
$(a + 3b) + i(3a + b) = -14 - 6i$

3. Equate real and imaginary parts and solve the resulting simultaneous equations.

Equating real and imaginary parts gives a system of equations:
$a + 3b = -14 \quad \text{--- (1)}$
$3a + b = -6 \quad \text{--- (2)}$

From (2), $b = -3a - 6$. Substitute this into (1):
$a + 3(-3a - 6) = -14$
$a - 9a - 18 = -14$
$-8a = 4 \implies a = -\frac{1}{2}$

Substitute $a$ back to find $b$:
$b = -3\left(-\frac{1}{2}\right) - 6 = -\frac{9}{2}$

So, $z = -\frac{1}{2} - \frac{9}{2}i$.

4. Find $w$ using the previously found value of $z^*$.

Since $z = -\frac{1}{2} - \frac{9}{2}i$, its conjugate is $z^* = -\frac{1}{2} + \frac{9}{2}i$.
Substitute $z^*$ back into the expression for $w$:
$w = -2 + 4i - \left(-\frac{1}{2} + \frac{9}{2}i\right)$
$w = \left(-2 + \frac{1}{2}\right) + i\left(4 - \frac{9}{2}\right)$
$w = -\frac{3}{2} - \frac{1}{2}i$