I can not figure out how to find $P(1.9\leq|X|)$ when $1.9$ is not given as any value of $x$. I did what I knew how to do, but I am not sure how to proceed from here. I have inserted my LaTex code below. The output can be seen 
Consider a random variable $X$ whose probability mass function is given by
$$p(x)=\begin{cases}p&\text{if $x=-3$}\\ 0.1&\text{if $x=-0.1$}\\ 0.3&\text{if $x=2p$}\\ p&\text{if $x=3$}\\ p^2&\text{if $x=4$}\\ 0&\text{if otherwise}\\ \end{cases} $$
What is $p$?
We know that the total probability must add up to 1.
Thus we get the following:$p^2+2p+0.4=1\Rightarrow p^2+2p-0.6\Rightarrow p=0.264911$
Thus the probability mass function is actually as follows:
$$p(x)= \begin{cases} 0.264911&\text{if $x=-3$}\\ 0.1&\text{if $x=-0.1$}\\ 0.3&\text{if $x=0.529822$}\\ 0.264911&\text{if $x=3$}\\ 0.070178&\text{if $x=4$}\\ 0&\text{if otherwise}\\ \end{cases} $$
Now we can make the following C.D.F:
$$\begin{array}{|l|l|l|l|l|l|}\hline x& -3 & -0.1 & 0.53 & & 4 \\ \hline P(X\leq x) & 0.265 & 0.365 & 0.665 & 0.93 & 1 \\ \hline \end{array}$$
Compute $P(1.9\leq|X|\leq 3)$
First we will calculate $P(|X|\leq 3)$
$ P(|X|\leq 3)=0.264911+0.3+0.1+0.264911 = 0.93$
What is $P(2x-3\leq4|X\geq 2.0)$
Compute $Var(X)$
Compute $E\left(F\left(p\right)\left(X\right)\right)$
Hint 1: $$|X|\le 3 \iff -3\le X\le 3;\\ 1.9\le |X| \iff X\le -1.9 \ \ \text{OR} \ \ X\ge 1.9.$$ Hint 2: $$P(1.9\le |X|\le 3)=P(1.9\le |X|)+P(|X|\le 3)-1.$$ Hint 3: $$P(2x-3\leq4|X\geq 2.0)=P(X\le 3.5|X\geq 2.0)=\frac{P(2.0\le X\le 3.5)}{P(X\ge 2.0)}.$$