Expectation of probability less than r-th sample in order statistics 
How can that be $r/(n+1)$? in my lecture notes, the distribution function of the r-th order statistic is 
but I can't figure out how to get (a) from this equation.
 A: Summary
This diagram displays the key idea.

Assume all $n+1$ variables are independent, have the same distribution, and have no chance of coinciding.  Then $X$ will be to the left of $X_{r:n}$ a proportion $r/(n+1)$ of the time.
The Problem
For (a) to be correct,  $X_1, \ldots, X_n$, and $X$ must be independent continuous random variables with the same distribution $F$.  We are asked to demonstrate this mysterious-looking equation:
$$\mathbb{E}\left(\Pr(X \le X_{r:n})\right) = \frac{r}{n+1}\tag{1}$$
and then, using this result, we may conclude that $X_{r:n}$ ought approximately equal $F^{-1}(r/(n+1))$.  It is this intended conclusion that helps us resolve the inherent ambiguity in $(1)$: after all, to what variable(s) does the expectation $\mathbb{E}$ apply and to what variables(s) does the probability $\Pr$ apply?  To find out, we should try to understand $(1)$ in terms of $F$.
Interpreting the Question
Recall the definition of the distribution function $F$: for any real number $x$,
$$F(x) = \Pr(X \le x).\tag{2}$$
No ambiguity there: $X$ is the random variable.  Comparing this to $(1)$ suggests we should (temporarily) view $X_{r:n}$ as just a number, because $X_{r:n}$ evidently plays the role of $x$ in $(2)$.  Thus the probability is taken over $X$, whereas the expectation applies to $X_{r:n}$.
Analysis
The difficulty with mixing up expectations and probabilities in $(1)$ is primarily technical, not conceptual, so let's take care of the technical issue right now.  The probability of an event $\Pr(X \le x)$ can always be rewritten as the expectation of the indicator variable
$$\mathcal{I}(X\le x) = \cases{\eqalign{1, &\ & X\le x \\ 0, &\ &\text{otherwise.}}}$$
This allows us to rewrite $(1)$ solely in terms of expectations as
$$\mathbb{E}\left(\Pr(X \le X_{r:n})\right) = \mathbb{E}_{X_{r:n}}\left(\mathbb{E}_X(\mathcal{I}(X \le X_{r:n})\right).$$
To be clear, the subscripts indicate which variables the expectations are being taken over.  However, as you well know (and is intuitively obvious), it doesn't matter, because this is a total expectation:
$$\mathbb{E}_{X_{r:n}}\left(\mathbb{E}_X(\mathcal{I}(X \le X_{r:n}))\right) = \mathbb{E}_{(X_{r:n}, X)}(\mathcal{I}(X \le X_{r:n}))\tag{3}.$$
We may reverse the process that took us from a probability to an indicator: this expectation $(3)$ is merely the probability
$$\mathbb{E}_{(X_{r:n}, X)}(\mathcal{I}(X \le X_{r:n})) = \Pr(X \le X_{r:n}) .$$
This probability is taken over the full $n+1$-variate distribution of the variables $(X_1, X_2, \ldots, X_n, X)$.
(There are other ways to analyze the problem, but this method avoids any use of conditional probabilities, which can be tricky.)
Solution
It rests on this simple, elegant observation:

$X \le X_{r:n}$ means $X$ is among the $r$ least values when all $n+1$ variables are lined up.$\tag{4}$

That is because, by definition, at least $r-1$ of the $X_1, X_2, \ldots, X_n$ are less than $X_{r:n}$ but there is zero chance that $X_{(r-1):n}$ equals $X_{r:n}$.  (This is one reason we must assume $F$ is continuous.)
Because all $n+1$ variables have the same distribution, are independent, and have zero chance of being tied, any one of these variables--including $X$ itself--has equal chances of appearing in any one of the $n+1$ positions when all $n+1$ are lined up from smallest to largest.  Given $n+1$ equally likely possibilities, of which $r$ correspond to the event in question (according to $(4)$), we may immediately conclude
$$\Pr(X \le X_{r:n}) = \frac{r}{n+1}.$$
