Sorry for the bad title, I am not sure how do I name it.
Find all the roots that satisfy $z^4$ $$z^4 =\frac 12 e^{-i{\frac π7}} $$ $$z^4 = \frac 12 e^{i{\frac {13\pi}7}} $$
Therefore, the roots are. $$ z_1 = \sqrt[4]{\frac12}e^{i{\frac {13\pi}{28}}}, z_2 = \sqrt[4]{\frac12}e^{i{\frac {27\pi}{28}}}, z_3 = \sqrt[4]{\frac12}e^{i{\frac {41\pi}{28}}}, z_4 = \sqrt[4]{\frac12}e^{i{\frac {55\pi}{28}}}$$
I'm puzzled based on what did the above solution changed from $z^4 =$ $\frac 12 e^{-i{\frac π7}}$ to $ \frac 12 e^{i{\frac {13\pi}7}}$
Because $\frac 12 e^{-i\frac {\pi}{7}} = \frac 12 e^{i\frac {13\pi}{7}}$. Any complex number, in polar form $r e^{i\theta}$, is also equal to $r e^{i(\theta + 2k\pi)}$ for all $k\in \mathbb{Z}$.
This because, for $k\in \mathbb{Z}$, $e^{i2k\pi} = 1$, so $$r e^{i(\theta + 2k\pi)} =r e^{i\theta}e^{i2k\pi} =r e^{i\theta}\cdot 1 = r e^{i\theta}.$$
In your question they have used this with $k=1,2,3,4$, to say
$$\frac{1}{2} e^{-i\frac {\pi}{7}} = \frac{1}{2} e^{i\frac {13\pi}{7}} = \frac{1}{2} e^{i\frac {27\pi}7} = \frac{1}{2} e^{i\frac {41\pi}{7}} = \frac{1}{2} e^{i\frac {55\pi}{7}}.$$
The four fourth-roots follow from that.