On the interval $(0,1)$ $9$ points are chosen at random.Let $X$ be the distance from $0$ to the medium between the chosen points. On the interval $(0,1)$ 9 points are chosen at random.- This most likely means uniformly, I doesn't say more than was is written, just the presumption of choosing these points is what comes to mind naturally reading it.
Let $X$ be the distance from $0$ to the medium between the chosen points(meaning the distance from $0$). Find the density function of $X$ and the expected value as well... To be honest I haven't made any tangible progress toward the answer.. 
 A: Let $W$ be the distance from $0$ to the fifth largest point. We want to find $F_W(w)$, the probability that $W \le w$. This is the probability that at least $5$ of our numbers are $\le w$. The density can then be obtained by differentiating.
The probability exactly $5$ are $\le w$ is $\binom{9}{5}w^5(1-w)^4$.
The probability exactly $6$ are $\le w$ is $\binom{9}{6}w^6 (1-w)^3$.
And so on.
Remarks: $1.$ We do not need the density function in order to find the mean of $W$: it is $\frac{1}{2}$ by symmetry.
$2.$ The expression for $F_W(w)$ in the answer above is somewhat messy, and then we need to differentiate. Here is an alternate way to obtain the density directly. The idea is useful for calculating the density of the order statistics for a sample of $n$ from a continuous distribution. 
We argue informally. The median lies between $w$ and $w+dw$ with probability about $f_W(w)\,dw$. We compute this probability another way. We want $1$ observation to be between $w$ and $w+dw$, and $4$ of the observations to lie below, and $4$ to lie above. The probability of $2$ or more observations in the interval is negligible, so our probability is approximately $\left(\binom{9}{1}dw\right)\binom{8}{4}w^4(1-w)^4)$. That yields $f_W(w)=\binom{9}{1}\binom{8}{4}w^4(1-w)^4$.
A: I'm assuming you mean median, not medium. Think of it this way, what're the odds that the median is $\leq M$ for some $M$? We'd have to pick at least 5 points below $M$ and the rest above $M$, for any choice of points. That is:
$$
\begin{split}
Pr(X \leq M) &= \binom{9}{5}M^5(1-M)^4 \\
&+ \binom{9}{6}M^6(1-M)^3 \\
&+ \binom{9}{7}M^7(1-M)^2 \\
&+ \binom{9}{8}M^8(1-M)^1 \\
&+  M^9
\end{split}$$
As a sanity check, $Pr(X \leq 1) = 1$, since the first four terms are $0$. To get the pdf, we just differentiate the above, coming up with:
$$PDF = 630M^4(1-M)^4$$
And then to find the expected value:
$$\begin{split}
E[X] &= \int_0^1630M^5(1-M)^4dM \\
&= \left.630(\frac{M^{10}}{10}-\frac{4M^9}{9}+\frac{3M^8}{4}-\frac{4M^7}{7}+\frac{M^6}{6})\right|_0^1 \\
&= \frac12
\end{split}$$
A: For the pdf, you have four numbers less than $x$ and four numbers greater than $x$, so the pdf is proportional to $x^4(1-x)^4$.  All you have to do is normalize it, so the integral from $0$ to $1$ is $1$.
