Finding Complex Number $z$ in $\frac{z+2i}{z-2i}=\frac{7-6i}{5}$ 
What I did: Cross Multiply, try to expand out the mod and args, but they all seem to lead to dead end (probably I am not seeing something)

 A: All your equations are correct but apparently your main difficulty was how
to find $z$ in the algebraic form $a+bi$ from the given equation, which you
have shown is equivalent to the linear equation in $z$
$$
2z-12-(24+6z)i=0.
$$
To solve this equation add the terms in $z$ and separately the independent
terms.
$$
\left( 2-6i\right) z-(12+24i)=0.
$$
Move the independent term to the RHS.
$$
\left( 2-6i\right) z=12+24i.
$$
To find $z$ divide both sides by the coefficient of $z$, multiply both
numerator and denominator by the conjugate of the denominator and simplify
$$ 
\begin{eqnarray*}
z &=&\frac{12+24i}{2-6i}=\frac{6+12i}{1-3i}=\frac{\left( 6+12i\right) \left(
1+3i\right) }{\left( 1-3i\right) \left( 1+3i\right) } \\
&=&\frac{-30+30i}{10}=-3+3i.
\end{eqnarray*}
$$
So $a=-3,b=3$. You can easily find the modulus of $z$ 
$$
\left\vert z\right\vert =\sqrt{a^{2}+b^{2}}=\sqrt{(-3)^{2}+3^{2}}=3\sqrt{2}.
$$
Its argument $\theta =\arg z$ is the angle in the 2nd quadrant$^1$ such that $\tan \theta =b/a$. The range of the arctangent function is $]−\pi/2, \pi[$. Then 
$$
\arg z=\pi+\arctan \left( \frac{b}{a}\right) =\pi+\arctan \left( \frac{3}{-3}\right)
=\pi+\arctan \left( -1\right) =\pi-\frac{1}{4}\pi=\frac{3}{4}\pi .
$$
As for $w$ you do not need to find its algebraic form. You can apply the
properties of the modulus and argument of a complex fraction. Since 
$$
\left\vert \frac{2z}{w}\right\vert =\frac{\left\vert 2z\right\vert }{
\left\vert w\right\vert }=\frac{2\left\vert z\right\vert }{\left\vert
w\right\vert }=\frac{6\sqrt{2}}{\left\vert w\right\vert }=3,
$$
solving for $\left\vert w\right\vert $ you get $\left\vert w\right\vert =
\frac{6\sqrt{2}}{3}=2\sqrt{2}$, while from the following equation 
$$
\arg \left( \frac{w}{z}\right) =\arg \left( w\right) -\arg \left( z\right)
=\arg \left( w\right) -\frac{3}{4}\pi =-\frac{1}{8}\pi 
$$
you find $\arg \left( w\right) =-\pi /8+3\pi /4=5\pi /8$.
$^1$ added and corrected
A: If you solve $(2 - 6i)z - (12 + 24i) = 0$ you should get that :
$z=-3+3i \Rightarrow |z|=\sqrt{(-3)^2+3^2}=3\cdot \sqrt{2}$ , and 
$\arg z= \arctan(\frac{3}{-3})+ \pi=\frac{7\pi}{4}$
I assume that you can do the rest of task... 
