= Solution
Normalization of the <probability density function> gives
$$
1=c\int_0^{\sqrt3}\frac{dx}{1+x^2}
=c\arctan(\sqrt3)
=c\frac\pi3,
$$
so
$$
\boxed{c=\frac3\pi}.
$$
The <expected value> is
$$
\mathbb EX
=\frac3\pi\int_0^{\sqrt3}\frac{x}{1+x^2}\,dx
=\boxed{\frac{3\log2}{\pi}}.
$$
The second moment is
$$
\mathbb EX^2
=\frac3\pi\int_0^{\sqrt3}\frac{x^2}{1+x^2}\,dx
=\frac{3\sqrt3}{\pi}-1.
$$
Therefore the <variance> is
$$
\boxed{
\operatorname{var}X
=\frac{3\sqrt3}{\pi}-1
-\frac{9(\log2)^2}{\pi^2}}.
$$
For the <median> $m$,
$$
\frac3\pi\arctan m=\frac12,
$$
so
$$
\boxed{m=\tan\frac\pi6=\frac1{\sqrt3}}.
$$
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