Consider
$$\frac{a + jb}{c + jd}$$
Now we can get rid of the awkward term on the bottom by multiplying $\frac{a + jb}{c + jd}$ by 1. and $\frac{c-jd}{c-jd}$ is 1
$$\frac{a + jb}{c + jd} \cdot \frac{c - jd}{c - jd} = \frac{a \cdot c - j^2b \cdot d + j b \cdot c - j a \cdot d}{c^2 - j^2 d^2} = \frac{a \cdot c + b \cdot d + j(b \cdot c - a \cdot d)}{c^2 +d^2}$$
This works because when you multiply any number by its complex conjugate the result is real.
So you multiply the denominator by its complex conjugate to make it real and you multiply the numerator by the same number so you have in effect multiplied the fraction by 1.
To solve your particular example I would first convert $8\angle-20$ to $a+jb$ form because its easier to add complex numbers in this form
$$\begin{align}
\frac{8 \angle -20^o}{(2+j)(3-4j)}+\frac{10}{-5+12j} & = \frac{7.518 - 2.736j}{(2+j)(3-4j)}+\frac{10}{-5+12j}\\
& = \frac{7.518 - 2.736j}{10 - 5j}+\frac{10}{-5+12j}\\
& = \frac{7.518 - 2.736j}{10 - 5j} \cdot \frac{10+5j}{10+5j}+\frac{10}{-5+12j}\cdot \frac{-5 -12j}{-5 -12j}\\
& = \frac{88.86 + 10.23j}{125}+\frac{-50 - 120j}{169}\\
& = 0.415 - 0.628j
\end{align}$$
Which if you want in polar form
$$
0.415 - 0.628j = \sqrt{0.415^2+0.628^2} \angle \arctan \left( \frac{0.628}{0.415} \right) = 0.567 \angle 56.54^o$$
$r = \sin(2\theta) = 2\sin\theta\cdot \cos\theta \to r^3 = 2(r\sin\theta)(r\cos\theta)$. Then use:
$x = r\cos\theta$, and $y = r\sin\theta$, and $r = \sqrt{x^2 + y^2}$ to finish.
Best Answer
First convert both the numbers into complex or rectangular forms. ($j$ is generally used instead of $i$ as $i$ is used for current in Physics and Electronics, if you're related to these)
Add both and convert the sum back into polar form