Q:

probability of 1st order statistics is smaller than $x$ and 2nd order statistics is larger than $x$

Suppose that there $X_1,X_2,\cdots,X_n$ are drawn independently drawn according to a CDF $F$ and pdf $f$. 
Let $X_{(k;n)}$ be the $k$-th order statistics. so that we have $X_{(k;n)}\leq X_{(k+1;n)}$. Let's denote the CDF of $X_{(k;n)}$ by $F_{(k;n)}$.
Then what would be the probability that $X_{(1;n)}\leq x\leq X_{(2;n)}$? for some parameter $x\in \mathbb R$

A:

Since it has a pdf, the distribution is continuous.  You're asking for the probability that exactly one of your sample is $\le x$, and by the binomial distribution this is $n F(x) (1-F(x))^{n-1}$.  

