$$f(x) = \begin{cases} 2k, & \text{if $x$ = -3} \\[2ex] 0.4, & \text{if $x$ = 0} \\[2ex] 0.3 & \text{if $x$ =3 } \\[2ex] k & \text{if $x$ = -4} \\[2ex] 0 & \text{elsewhere} \end{cases}$$
$(1)$ Find the value of $k$;
$(2)$ Find the cumulative function $F(x)$.
$(3$) Find the variance of X.
$1.$ The first part one must use the formula
$F(x)= P(X \le x) = \sum_{t \le x } f(t)$
$\Rightarrow \sum_{x=1} f(x) =1$
$2k+0.4+0.3+k =1$
$k=.10$
$2.$ For this part finding the cumulative function is giving me some difficulties.
$$F(x) = \begin{cases} 0, & \text{if $x$ < -4} \\[2ex] 0.1, & \text{if -4 $ \le x <-3$ } \\[2ex] 0.4 & \text{if -3$<x \le$ 0 } \\[2ex] 0.3 & \text{if $ 0<x \le 3$ } \\[2ex] 1 & \text{x $\ge 3$} \end{cases}$$
This is from my interpretation it may not be correct.
$3.$ To Find the variance one uses
$\sigma^2_{x} = E(x^2) - E(x)^2 $
$E(X) = 2(.1)+.4+.3+.1 =1$
$E(X^2) = (2(.1)+.4+.3+.1)^2 =1$
$\sigma^2= (1)-(1)^2 =0$
Is this the correct way of finding $\mu_{1} \text{ and } \mu_{2}$?
Your CDF is incorrect, though you have written the correct formula.
Sum probabilities less than or equal to a given $x_i$ to yield $F(x_i)$.
$$ F(X) = \begin{cases} 0.1 \quad x = -4 \\ 0.3 \quad x = -3\\ 0.7 \quad x = 0 \\ 1 \quad x = 3 \end{cases}$$
Your mean is also false.
$$\mu = \sum p_ix_i =- 0.1$$
Note that your variance is also false. A random variable should have 0 variance of it is constant.