Since you're working with a discrete random variable, the probability of X being lower than b equals the probability of X being equal to or lower than the number before b.
Imagine we're working with a fair dice:
$P(2\leq X < 5) = P(2\leq X \leq 4)$. Sounds quite fair, doesn't it?
I think you can answer your question yourself now :)
These types of distributions are common in actuarial contexts, such as when we are considering an insurer's incurred loss on a policy that has a deductible. For instance, if ground-up losses (before deductible) on an automobile policy are exponentially distributed with a mean of \$1500, but the policy has a deductible of \$500, the insurer is only responsible for ground-up loss minus the deductible: if the insured experiences a loss that does not exceed the deductible--for instance, they get a small dent or a broken headlight--they don't file a claim because the cost of repair does not exceed the deductible, and the insurer's liability is zero. But if the policyholder has a major collision, the insured will file a claim and the insurer will pay the cost of repair minus the deductible.
So, we see that the insurer's liability is modeled by a mixed random variable $$Y = \begin{cases} 0, & X \le 500, \\ X - 500, & X > 500, \end{cases}$$ where $$f_X(x) = \frac{1}{1500}e^{-x/1500}, \quad x > 0$$ is the density for $X$. The probability distribution for $Y$ has a discrete portion since $$\Pr[Y = 0] = \Pr[X \le 500] = F_X(500) = 1 - e^{-1/3}.$$ The continuous portion of the distribution for $Y$ can be understood by noting that if $X > 500$, then $$F_Y(y) = \Pr[Y \le y] = \Pr[X - 500 < y] = \Pr[X \le y + 500] = F_X(y+500) = 1 - e^{-(y+500)/1500}.$$ Thus $$F_Y(y) = \begin{cases} 0, & y < 0 \\ 1 - e^{-(y+500)/1500}, & y \ge 0.\end{cases}$$ Note this function is not continuous at $y = 0$. Differentiation gives the continuous portion--i.e., the density--of the distribution for $Y$, but to specify things completely, we must write: $$\begin{align*} \Pr[Y = 0] &= 1 - e^{-1/3}, \\ f_Y(y) &= \frac{1}{1500}e^{-(y+500)/1500}, \quad y > 0. \end{align*}$$ I leave as an exercise for the reader to consider how we might compute the expected value and variance of $Y$, as well as how we might model a situation where the deductible is not ordinary as in this example, but a franchise deductible--i.e., the deductible is waived if a loss exceeding that amount occurs. It is also interesting to consider the case of policy limits--i.e., what happens to $Y$ if the insurer only pays up to a certain amount, say $L = 100000$.
Best Answer
It is neither discrete nor continuous. $P(X=1)=\frac 12 -\frac1 4=\frac 1 4$ (the jump at $x=1$) etc.
For any random variable $X$ we have $P(X=x)=F_X(x)-F_X(x-)$ where $F_X(x-)$ is the left hand limit of $F_X$ at $x$. In this case the right hand limit of $F$ at $1$ is $\frac 1 2$ and the left hand limit is $\frac 1 4$ so the difference is $\frac 1 4$.