Is it possible to extra/interpolate probabilities? Is it possible to extra/interpolate probabilities ? 
Lets say: $\qquad$ $P(X\geq3)=y\qquad$ and $\qquad P(X\leq5)=z$ $\qquad$($y,z$ are already known)
Can I calculate $P(X=4)$ and $P(X=6)$?
Bonus question: 
Is it possible to calculate $F(i)=P(X\leq i)$ faster than adding $f(i)+f(i-1)+...+f(0)$?
 A: No, it is not possible.

In special cases, if your probabilities $y$ and $z$ are equal to $0$ or $1$, it is possible to infer some facts. 
For example, if we know that $P(X\geq 3) = 0$ then we know that $P(X =4) = 0$ and $P(X = 6)= 0$. 
Or, if you know that $P(X \leq 5) = 1$ then we know that $P(X = 6) = 0$. 
Or, if you know that $P(X \leq 5) = 0$ then we know that $P(X = 4) = 0$. 
A: OP states in one of the comments: 

In my practice Problems I have one like this: In 10 minutes there is a probability of 0.8 that at least 3 people are passing by, the probability that 5 people at most are passing by is 0.4. I need to find the probability of 3,4 and 6 people passing by.

I am thinking there is likely a typo in the last part. That is, the textbook author might have meant the setup as this (converted symbolically): Given $P(X\geq 3)=0.8$ and $P(X\leq 5)=0.4$, find $P(3 \leq X \leq 5)$.
Making the leap that this is the intended question (considering an introductory probability question), then this becomes rather elementary:
\begin{align}
P(3 \leq X \leq 5) &= P(X \leq 5)-P(X \leq 2)\\
&= P(X \leq 5)-\left\{1-P(X \geq 3)\right\}\\
&= 0.4 - \left\{1-0.8\right\} = 0.4-0.2=0.2
\end{align}
