The cumulative probability distribution function $F_X(x)$ tells us how much
probability mass there is to the left of $x$ or at $x$ for each $x$
on the real line. (The choice of notation, though almost universally used
is truly dreadful for use in a classroom setting! How on earth does one read out aloud $F_X(x)$ or $P\{X\leq x\}$? F-sub-big X of little x? probability that
random variable $X$ is no larger than lower-case x?) Formally, the value of $F_X(x)$ is just $P\{X \leq x\}$.
As Glen_b's comment says, you really should start by sketching the function
$F_X(x)$ at the very least.
When $X$ is a discrete random variable taking on values $x_1, x_2, \ldots$
with probabilities $p_1, p_2, \ldots $ respectively, a little thought
(instead of rote memorization of the definition) reveals that $F_X(x)$
must be what can be described as a staircase function, increasing from
$0$ to $1$ as $x$ increases, with steps of heights $p_1, p_2, \ldots$
at points $x_1, x_2, \ldots$ etc. The function is discontinuous
at each $x_i$, and is constant in each interval $[x_i, x_{i+1})$ (please
be sure to note the $[$ and $)$ in the description of the intervals).
Note that $F_X(x_i)$ includes $p_i$ so that the value of $F_X(x)$ at
the point $x=x_i$ (where the function is discontinuous) is the
value on the right. Since you are studying from a text intended for
engineers, you might find this written as $F_X(x) = F_X(x^+)$.
Thus,
$$F_X(x) = P\{X \leq x\} = F_X(x^+) ~ \text{and} ~ P\{X < x\} = F_X(x^-).$$
In fact, for any random variable (not necessarily a discrete random variable
or an integer-valued random variable as in Rusan's answer) and for
any real numbers $a$ and $b$ such that $a \leq b$,
$$\begin{align}
P\{a < X \leq b\} &= F_X(b^+) - F_X(a^+) = F_X(b)-F_X(a),\tag{1}\\
P\{a \leq X \leq b\} &= F_X(b^+) - F_X(a^-) = F_X(b) - F_X(a^-),\tag{2}\\
P\{a \leq X < b\} &= F_X(b^-) - F_X(a^-) = F_X(b^-) - F_X(a^-),\tag{3}\\
P\{a < X < b\} &= F_X(b^-) - F_X(a^+) = F_X(b^-)-F_X(a).\tag{4}
\end{align}$$
For the special case when $b = a$, $(2)$ above becomes
$$P\{X=a\} = F_X(a^+)-F_X(a^-),$$ that is, $P\{X=a\}$ is the jump
(if any) in the value of $F_X(x)$ at $x=a$. If $F_X(x)$ is continuous
at $x=a$, then $P\{X=a\}=0$.
With this as prologue, note that your given $F_X(x)$ is a staircase
function with jumps of $\frac 14, \frac 12, \frac 14$ at $x=-10, 30, 50$
respectively; that is, $X$ takes on values $-10, 30, 50$ with
probabilities $\frac 14, \frac 12, \frac 14$ respectively, and
once you have that, the answers to the questions asked are easy
to compute directly, or, if you prefer to read the $F_X(0^-)$
etc off the graph that you have drawn as you apply $(1)$-$(4)$,
that is fine too.
Let's start with the definitions.
The cumulative distribution function, or CDF, of a random variable $X$ is the function $F_X:\mathbb{R}\to\mathbb{R}$ defined by
$$F_X(x) = \Pr(X \le x).$$
A discrete random variable $X$ which takes on the values $(x_1, x_2, x_3, \ldots)$ with probabilities $(p_1, p_2, p_3, \ldots)$, respectively, therefore has distribution function
$$F_X(x) = \sum_{i\,|\, x_i \le x} p_i.$$
This can be written conveniently in terms of the Heaviside function
$$\theta(x) = \begin{array}{ll} \left\{
\begin{array}{ll}
0 & x \lt 0 \\
1 & x\ge 0
\end{array}\right.
\end{array}$$
via
$$F_X(x) = \sum_i p_i \theta(x-x_i).$$
To say that a random variable $X$ has a uniform distribution supported on $[a,b]$, written $X\sim U(a,b)$, means that its CDF rises linearly from $0$ at $x=a$ to $1$ at $x=b$. There is no probability that $X$ can lie outside the interval $[a,b]$, implying that $F_X(x)=0$ for $x\lt a$ and $F_X(x) = 1$ for $x\gt b$. To describe such a piecewise linear function, introduce the function
$$\Theta(x) = x^{+} = \max(0, x).$$
This function is $0$ for $x\lt 0$ and otherwise equals $x$ for $x\ge 0$. We may write
$$F_X(x) = \Theta(z) - \Theta(z-1)$$
where $z = (x-a)/(b-a)$.
A mixture of distributions $F_1, F_2, F_3, \ldots$ with weights $\omega_1, \omega_2, \omega_3, \ldots$, respectively, will be a distribution when the weights are non-negative and sum to unity. In that case the distribution function of the mixture is
$$F(x) = \sum_i \omega_i F_i(x).$$
The last definition enables us to write the answer immediately as
$$F(x) = \frac{1}{3}F_1(x) + \frac{1}{3} F_2(x) + \frac{1}{3} F_3(x)$$
where the first distribution $F_1$ is $U(1,2)$, the second $F_2$ is $U(2,4)$, and the third $F_3$ is discrete. Applying definitions (2) and (3) as appropriate enables us immediately to write
$$F_1(x) = \Theta\left(\frac{x-1}{2-1}\right) - \Theta\left(\frac{x-1}{2-1} - 1\right),$$
$$F_2(x) = \Theta\left(\frac{x-2}{4-2}\right) - \Theta\left(\frac{x-2}{4-2} - 1\right),$$
and
$$F_3(x) = 0.4 \theta(x - 2) + 0.6 \theta(x - 3).$$
Here are their graphs.
Plugging them into the formula for $F$ gives a solution. The remaining issue concerns how to simplify it. Arguably, one would usually not want to "simplify" this expression, because that would only obscure what it means and how to interpret it. One valid reason for a simplification would be to improve computation, but the separate computation of the $F_i$ is already so simple and fast that even this reason seems like a poor one.
However, it may be of interest to understand $F$ qualitatively. To this end, note that
$F_1$ and $F_2$ are piecewise linear and continuous, but have "kinks" where a slope is not defined. (The slope, where defined, gives a value for a probability density function and therefore can be of some interest.)
$F_3$ is piecewise constant, but contains jumps.
Because the mixture is a linear combination of the $F_i$ and linear combinations preserve linearity, we know immediately that $F$ is piecewise linear but may have kinks and jumps. Where are these interesting values located? They can be only where they might occur in the $F_i$. Looking at the plots (or the formulas, for those who prefer algebraic reasoning to geometric), it should be clear that a $U(a,b)$ distribution has kinks only at $\{a,b\}$--which is $\{1,2\}$ for $F_1$ and $\{2,4\}$ for $F_2$--and a discrete distribution has jumps only at its support $\{x_i\}$, which for $F_3$ is $\{2,3\}$. Therefore:
$F$ can have jumps only at $\{2,3\}$.
$F$ can have kinks only at $\{1,2,4\}$.
$F$ must be linear in between any kinks or jumps.
Thus, $F$ must have a piecewise linear definition broken down on the intervals $x\lt 1$, $1\le x \lt 2$, $2 \le x \lt 3$, $3 \le x \lt 4$, and $x\ge 4$. To write this down explicitly--which is rarely needed--use your favorite methods to work out the formulas of linear functions.
$$F(x) = \begin{array}{cc}
\left\{
\begin{array}{cc}
0 & x \lt 1 \\
\frac{1}{3}(x-1) & 1\le x\lt2 \\
\frac{1}{30} (5x+4) & 2\leq x\lt3 \\
\frac{1}{6}(x+2) & 3\leq x\lt 4 \\
1 & x\ge 4. \\
\end{array}\right.
\end{array}$$
This answer uses the word "immediately" three times at junctures where no thought was required: only rote substitution of formulas into other formulas or application of a definition was needed. These applications are fundamentally uninteresting. What is of interest, and worth remembering from doing such an exercise, is the idea that certain qualitative aspects of mixture distribution can easily be inferred from qualitative properties of their components. This question focuses on reasoning about properties of continuity (absence of jumps) and differentiability (absence of jumps or kinks).
Best Answer
Because if $x \leq y$, then if $X \leq x$, it follows that $X \leq y$. Therefore, $P(X \leq x) \leq P(X \leq y)$.
More generally, probabilities are monotone in the sense that if $A$ and $B$ are events and $A \subseteq B$, then $P(A) \leq P(B)$. This follows from writing $B$ as the disjoint union of $A$ and $B \setminus A$, whence by the probability axioms $P(B) = P(A) + P(B \setminus A) \geq P(A)$ (since $P(B \setminus A) \geq 0$).
In the case of cumulative distribution functions with $x \leq y$, take $A = \{X \leq x\}$ and $B = \{X \leq y\}$.