In a school with a large number of students, $p\%$, where $p > 40$, of its students study H2 Biology.

1. Define the binomial random variable for the number of students studying H2 Biology.

Let $X$ be the number of students studying H2 Biology in the sample of 40.
$X \sim \text{B}\left(40, \frac{p}{100}\right)$

2. Formulate the equation based on the given probability.

We are given that the probability of no less than 9 but no more than 20 students studying H2 Biology is $0.25$.
$P(9 \le X \le 20) = 0.25$
$P(X \le 20) - P(X \le 8) = 0.25$

3. Solve the equation for $p$ using a graphing calculator.

Using a graphing calculator, we can graph $Y_1 = P(X \le 20) - P(X \le 8) - 0.25$ where the probability of success is $x = \frac{p}{100}$.
We find the root for $x > 0.4$ (since it is given that $p > 40$):
$x \approx 0.5653$
$\frac{p}{100} = 0.5653$
$p = 56.5 \text{ (to 3 s.f.)}$

1. Find the probability for the specific sequence of the first three interviewees.

Let $S$ denote a student who studies H2 Biology, and $S'$ denote a student who does not.
Given $p = 60$, the probability of a randomly chosen student studying H2 Biology is $P(S) = 0.6$, and $P(S') = 0.4$.
For the 3rd interviewee to be the 1st student studying H2 Biology, the 1st and 2nd interviewees must be $S'$, and the 3rd must be $S$.
$\text{Probability} = 0.4 \times 0.4 \times 0.6 = 0.096$

2. Find the probability for the 4th to 10th interviewees satisfying the conditions.

For the 10th interviewee to be the 4th student studying H2 Biology, there must be exactly 2 students studying H2 Biology among the 6 interviewees from the 4th to 9th positions, and the 10th interviewee must be $S$.
$\text{Probability} = \binom{6}{2} (0.6)^2 (0.4)^4 \times 0.6$
$= 15 \times 0.36 \times 0.0256 \times 0.6 = 0.082944$

3. Calculate the total probability of the entire sequence.

Since the selections of interviewees are independent, we multiply the probabilities of the required sequences for the first 10 interviewees. The outcomes of the 11th and 12th interviewees do not affect this condition.
$\text{Required probability} = 0.096 \times 0.082944$
$= 0.007962624$
$\approx 0.00796 \text{ (to 3 s.f.)}$