You noticed that $1+i=\sqrt{2}e^{\pi i/4}$. That implies that $8\sqrt{2}(1+i)=8\cdot\sqrt{2}\cdot\sqrt{2}\cdot e^{\pi i/4}=16e^{\pi i/4}.$ If $z=re^{i\theta}$, where $r>0$ and $\theta\in[0,2\pi)$, has the property that $z^4=8\sqrt{2}(1+i)$, then we must have that $z^4=(re^{i\theta})^4=r^4e^{4i\theta}=16e^{\pi i/4}.$ Because $r$ is a non-negative real number, we know that $r^4=16$ implies that $r=2$.
However, solving for $\theta$ is slightly trickier because angles "wrap around" after $2\pi$. To be precise, $e^{4i\theta}=e^{\pi i/4}$ implies that $4i\theta=\frac{\pi i}{4}+2\pi ik\text{ for some }k\in\mathbb{Z}$
and therefore, $\theta\in[0,2\pi)$ is one of the four values $\frac{\pi }{16}+k\frac{\pi}{2}$ where $k\in\{0,1,2,3\}$.