Understanding CDF and its relationship with Probability of r.v. 
I attempted to solve this problem myself. I am trying to solve the problem by applying the fundamental formula of CDF. I.e. $P(X≤x) = F_X(x)$. 
Let us see the problem (b).
(i) $P(X \le \frac{1}{4})$
$\Rightarrow F(\frac{1}{4})$  (according to definition)
$\Rightarrow \frac{1}{4} + \frac{1}{2}$
$\Rightarrow \frac{3}{4}$
(ii) $P(0 < X \le \frac{1}{4})$
$\Rightarrow P(X \le \frac{1}{4}) - P(X \le 0)$ (coz, it excludes $0$)
$\Rightarrow F(\frac{1}{4}) - F(0)$    (according to definition)
$\Rightarrow \frac{3}{4} - \frac{1}{2}$
$\Rightarrow \frac{1}{4}$ 
(iii) $P(X=0)$
$\Rightarrow P(X \le 0) - P(?)$ (couldn't figure out! kindly, help.)
(iv) $P(0 \le X \le \frac{1}{4})$
$\Rightarrow P(X \le \frac{1}{4}) - P(X \le 0) + P(0)$ (am I correct?)
(i) and (ii) are okay. But, I am not sure about (iii), and (iv).
Also, (iii) uses a peculiar notation: $F(0^-)$.

Not quite sure what you are asking since the answers are in the picture. Here is some reasoning behind iii) and iv).
iii) is given by $P(x\leq 0) - P(x<0)$ since we want the probability that $X=0$. Since $F(x)$ is defined to be $0$ for $x<0$ this gives $\frac{1}{2} - 0 = \frac{1}{2}$.
iv) We want the probability of $0\leq X\leq \frac{1}{4}$. If we take $F(\frac{1}{4}) - F(0)$ only then we leave out the probability that $X = \frac{1}{2}$ so this is added back in by the result in iii).
Update: 
$F(0^-)$ simply means $P(X<0)$. 
Your answer to (iv) is correct.
