Trying to solve discrete probability exercises I always don't know which formula I should use to calculate $P(A \cap B)$. I have 3 choices :
1) $P(A) \cdot P(B)$
2) $P(A|B) \cdot P(B)$ (from conditional probability formula)
3) Do some calculations in order to use DeMorgan's laws
I don't know when use 1) instead of 2) or 3) and viceversa. For example : The probability that a student, randomly chosen among the undergraduates in biology, after the first exam session doesn't pass the chemistry test is 0.4; the probability the student is from a scientific high school is 0.3, the probability that a student is from a scientific high school and she passed the test is 0.25. What is the probability that a student is not from a scientific high school and she didn't pass the chemistry test?
There's no magic rule when to use one or the other... you use whichever is most helpful, which you'll get a sense for as you gain more experience with this type of problem.
(Other than, as BruceET pointed out, that the first formula is valid if and only if $A$ and $B$ are independent.)
In your problem, we aren't given independence, and we don't have the probability of one event given another, but we do have the probability of some negations, so De Morgan looks like the best bet. Let $A$ be "the student is not from a scientific high school," and $B$ "the student didn't pass the test." We want $P(A \cap B)$. We have $P(B) = 0.4$, $P(\neg A) = 0.3$, and $P(\neg A \cap \neg B) = 0.25$. Looking at the quantities we know, it seems most promising to use $$\begin{align} P(A \cap B) &= 1 - P(\neg A \cup \neg B) \\&= 1 - P(\neg B) - P(\neg A) + P(\neg A \cap \neg B) \\&= P(B) - P(\neg A) + P(\neg A \cap \neg B) \\&= 0.4 - 0.3 + 0.25 \\&=0.35 \end{align}$$