13. Let $|z|=1$, find $f(z)=\max \left\{\frac{|1+z|}{2}, \sqrt{\frac{\left|1+z^{2}\right|}{2}}\right\}$ of the extremum.

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The translation is provided as requested, maintaining the original format and line breaks.
13. Let $z=\cos \theta+i \sin \theta$

So $\frac{|1+z|}{2}=\frac{\sqrt{(1+\cos \theta)^{2}+\sin ^{2} \theta}}{2}=\frac{\sqrt{2+2 \cos \theta}}{2}=\frac{\sqrt{4 \cos ^{2} \frac{\theta}{2}}}{2}=\left|\cos \frac{\theta}{2}\right|$
$$
\begin{aligned}
\sqrt{\frac{\left|1+z^{2}\right|}{2}} & =\sqrt{\frac{|1+\cos 2 \theta+i \sin 2 \theta|}{2}}=\sqrt{\frac{\left|2 \cos ^{2} \theta+2 i \sin \theta \cos \theta\right|}{2}}=\sqrt{|\cos \theta|} \\
& =\sqrt{\left|2 \cos ^{2} \frac{\theta}{2}-1\right|}
\end{aligned}
$$

Let $\frac{|1+z|}{2}=\left|\cos \frac{\theta}{2}\right|=x$, then $y=\sqrt{\frac{\left|1+z^{2}\right|}{2}}=\sqrt{\left|2 x^{2}-1\right|}(x>0, y>0)$
So $y^{2}=\left|2 x^{2}-1\right|$
(1) $2 x^{2}-1 \geqslant 0$
$3(\max \{x, y\})^{2} \geqslant 2 x^{2}+y^{2} \geqslant 2 x^{2}-y^{2}=1$
(2) $2 x^{2}-1 \leqslant 0 \quad 3(\max \{x, y\})^{2} \geqslant 2 x^{2}+y^{2}=1$

So $\max \{x, y\} \geqslant \frac{\sqrt{3}}{3}$, and when $\cos \frac{\theta}{2}=\frac{\sqrt{3}}{3}$, the equality holds.
Clearly, the maximum value of $\max \{x, y\}$ is 1, so the extreme values are $1, \frac{\sqrt{3}}{3}$.