Expected values of a sorted sequence If I have $n$ random variables uniformly distributed on [0,1]. What is
the expectance value kth highest value? In other words, what is the
expected value of sorting the random variables?
For $n=2$ the expectance for the highest value is $2/3$ while the second highest is $1/3$.

You're looking for the term "order statistic". If $X_1, \ldots, X_n$ are i.i.d. random variables, then the order statistic $X_{(1)}, \ldots, X_{(n)}$ of those random variables is a reordering of the values that satisfies $X_{(1)} \le \ldots \le X_{(n)}$. 
In the special case of the uniform distribution on $[0, 1]$, the distribution of the $k$-th order statistic $X_{(k)}$ is the beta-distribution $B(k, n + 1 - k)$. This means in particular that $E[X_{(k)}] = \frac{k}{n + 1}$.
You can calculate the density of the $k$-th order statistic in terms of the CDF and the density of the initial distribution via this formula. To prove it, observe that
$$P(X_{(k)} \le x) = \sum \limits_{(i_1, \ldots, i_n)} P(X_1 \in A_{i_1}) \ldots P(X_n \in A_{i_n})= \sum \limits_{j = k}^n \binom{n}{j} P(X_1 \le x)^j P(X_1 > x)^{n - j}$$
where $A_0 = (x, \infty)$, $A_1 = (-\infty, x]$ and the sum goes over all tuples $(i_1, \ldots, i_n) \in \{0, 1\}^n$ with $\sum \limits_{j = 1}^n i_j \ge k$. Differentiating the term yields the desired density.
