Probability of non inclusive range So I know to get a probability like $P(2\leq X\leq 4)$, you simply do $P(X\leq4) - P(X\leq1)$, but when there is a question like $P(2<X<4)$ what am I supposed to do? 
Not just limited to in between two values, I also don't know what to do if it's just $P(X<2)$, so far all our examples have been greater/less than or equal to, for $P(X\leq 2)$ you just do the cumulative distribution up to $2$ but what do I do if that $2$ is not included in the range?
 A: If $X$ is a continuous random variable then $\mathbb{P}(X \leq c) = \mathbb{P}(X < c)$, for $c$ some constant. This is because the cumulative probability is given by the integral, letting $f_X$ be the distribution function of $X$,
\begin{equation*}
 \mathbb{P}(X \leq c) = F_x(c) = \int^c_{-\infty} f_X(t)\,dt
\end{equation*}
If you're familiar with integral calculus then it should be clear why there's no difference between integrating over the interval $(-\infty, c]$ and $(-\infty, c)$. 
If $X$ is a discrete random variable then we can write our cumulative probability as (suppose $X$ can't be negative for simplicity)
\begin{align*}
  \mathbb{P}(X \leq c) &= \mathbb{P}(X = 0 \text{ or } X = 1 \text{ or } X = 2 ~\cdots~ X = c) \\
&= \mathbb{P}(X = 0) + \mathbb{P}(X = 1) + \mathbb{P}(X = 2) + \cdots + \mathbb{P}(X = c)
\end{align*}
whereas the probability $\mathbb{P}(X < c)$ is
\begin{align*}
  \mathbb{P}(X < c) &= \mathbb{P}(X = 0 \text{ or } X = 1 \text{ or } X = 2 ~\cdots~ X = c - 1) \\
&= \mathbb{P}(X = 0) + \mathbb{P}(X = 1) + \mathbb{P}(X = 2) + \cdots + \mathbb{P}(X = c - 1)
\end{align*}
So, supposing $X$ is strictly positive for simplicity,
\begin{align*}
  \mathbb{P}(X \leq 2) &= \mathbb{P}(X = 0) + \mathbb{P}(X = 1) + \mathbb{P}(X = 2) \\
    \mathbb{P}(X < 2) &= \mathbb{P}(X = 0) + \mathbb{P}(X = 1)
\end{align*}
A: The complement of $\{X \geq 2\}$ is $\{X < 2\}$, e.g. we have $\Pr\{X \geq 2\} = 1 - \Pr\{X < 2\}$
The first example you shown is imprecise. The general version should be
$$ \Pr\{2 \leq X \leq 4\} = \Pr\{X \leq 4\} - \Pr\{X < 2\}$$
Only, when you are given that $\Pr\{1 < X < 2\} = 0$, e.g. $X$ is discrete with integral support, then you have your claim.
In your question $\Pr\{X > 2\} = 1 - \Pr\{X \leq 2\} = 1 - F_X(2)$ so it can be expressed in terms of the CDF $F$ conveniently. 
If you want to express, say $\Pr\{X < 2\} = \Pr\{X \leq 2\} - \Pr\{X = 2\}$ in terms of the CDF, then it depends on whether it has a mass on $2$. If it has no mass on $2$, then it is simply $F_X(2)$; otherwise you need to subtract from that. Or equivalently 
$$\Pr\{X < 2\} = \lim_{x\to 2^-}F_X(x)$$
