Given the CDF of $X$, find the distribution of $X$ and $P(1 < X < 5)$ 
I know I have to find the distribution of X and then sum the probabilities of the values that apply the rule, but I just can't figure out how to do it.

Try to rewrite the probability of the event $a<X<b$.
We could consider the probability of $X$ being less than $b$ and then subtract the probability of $X$ being less than $a$. So at first, we are taking all the values of $X$ such that $X<5$, then we are excluding the values which are less than $1$. So $P(1<X<5)=P(X<5)-P(X<1)$.
By definition
$$F_X(x)=P(X\leq x)\quad\text{ and }\quad P(X<a)=\lim_{n\to \infty}F_X\left(a-\frac{1}{n}\right)$$
So you can write
\begin{eqnarray*}
P(1<X<5)&=&P(X<5)-P(X<1)\\
&=&\lim_{n\to \infty}F_X\left(5-\frac{1}{n}\right)-\lim_{n\to \infty}F_X\left(1-\frac{1}{n}\right)\\
&=&\lim_{n\to \infty}\frac{5-\frac{1}{n}-1}{6}-\lim_{n\to \infty}F_X\frac{1-\frac{1}{n}}{4}\\
&=&\frac{2}{3}-\frac{1}{4}=\frac{5}{12}
\end{eqnarray*}
So you don't necessary need to use the PDF.
For a geometric intuition here is the plot of your CDF and its values in $1$ and $5$.
