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
$$
0.415 - 0.628j = \sqrt{0.415^2+0.628^2} \angle \arctan \left( \frac{0.628}{0.415} \right) = 0.567 \angle 56.54^o$$
Let $a,b\in\mathbb{R}$ so that $$\sqrt{i+1} = a+bi$$
$$ i+1 = a^2 -b^2 +2abi $$
Equating real and imaginary parts, we have
$$2ab = 1$$
$$a^2 -b^2 = 1$$
Now we solve for $(a,b)$.
$$
\begin{align*}
b &= \frac{1}{2a}\\\\
\implies \,\,\, a^2 - \left(\frac{1}{2a}\right)^2 &= 1 \\\\
a^2 &= 1 + \frac{1}{4a^2}\\\\
4a^4 &= 4a^2 + 1\\\\
4a^4 - 4a^2 -1 &= 0 \\\\
\end{align*}
$$
This is a quadratic in $a^2$ (it's also a quadratic in $2a^2$, if you prefer!), so we use the quadratic formula:
$$a^2 = \frac{4 \pm \sqrt{16-4(4)(-1)}}{2(4)}$$
$$a^2 = \frac{1 \pm \sqrt{2}}{2}$$
Here we note that $a$ is real, so $a^2>0$, and we discard the negative case:
$$a^2 = \frac{1 + \sqrt{2}}{2}$$
$$a = \pm \sqrt{\frac{1 + \sqrt{2}}{2}}$$
$$ b = \frac{1}{2a} = \pm \sqrt{\frac{\sqrt{2}-1}{2}}$$
This gives what you can call the principal root:
$$\sqrt{i+1} = \sqrt{\frac{1 + \sqrt{2}}{2}} + i\sqrt{\frac{\sqrt{2}-1}{2}} $$
As well as the negation of it:
$$-\sqrt{i+1} = -\sqrt{\frac{1 + \sqrt{2}}{2}} + i\left(-\sqrt{\frac{\sqrt{2}-1}{2}}\right) $$
Finally, substituting either of these into your expression $$z=2i \pm \sqrt{i+1}$$ will give you $\text{Re}(z)$ and $\text{Im}(z)$.
At that point, as you noted in your question, conversion to polar coordinates is straightforward.
Best Answer
It depends on what is wanted. If you wat a polar form, then $cis(\pi/6)$ would be a correct answer. So are $cis(13\pi/6)$ and $cis(-11\pi/6)$. If instead, all polar forms must be rendered, then you would put $cis(\pi/6+2k\pi), k\in\mathbb{Z}$.