Suppose that $\alpha$ is a real number and $a_1<a_2<.....$ is a strictly increasing sequence of natural numbers such that for each natural number $n$ we have $a_n\le n^{\alpha}$. We call the prime number $q$ golden if there exists a natural number $m$ such that $q|a_m$. Suppose that $q_1<q_2<q_3<.....$ are all the golden prime numbers of the sequence $\{a_n\}$.

[b]a)[/b] Prove that if $\alpha=1.5$, then $q_n\le 1390^n$. Can you find a better bound for $q_n$?

[b]b)[/b] Prove that if $\alpha=2.4$, then $q_n\le 1390^{2n}$. Can you find a better bound for $q_n$?

[i]part [b]a[/b] proposed by mahyar sefidgaran by an idea of this question that the $n$th prime number is less than $2^{2n-2}$
part [b]b[/b] proposed by mostafa einollah zade[/i]
We are given a strictly increasing sequence of natural numbers \( \{a_n\} \) such that \( a_n \leq n^\alpha \) for each natural number \( n \). We need to prove bounds on the golden prime numbers \( q_n \) for different values of \( \alpha \).

### Part (a)
We need to prove that if \( \alpha = 1.5 \), then \( q_n \leq 1390^n \).

1. **Assume for contradiction**: Suppose there exists an \( n \) such that \( q_n > 35^n \). Let \( n \) be the minimal such index. Let \( r \) be the minimal index such that \( q_n \mid a_r \). Then, we have:
   \[
   r > 35^{\frac{2}{3}n} \quad (\clubsuit)
   \]

2. **Prime factors in the set \( P \)**: All of \( \{a_1, a_2, \ldots, a_{r-1}\} \) have prime factors in the set \( \{q_1, q_2, \ldots, q_{n-1}\} \). Call this set of primes \( P \).

3. **Sum of reciprocals**: We have:
   \[
   \sum_{k=1}^{r-1} \frac{1}{a_k^{\frac{1}{3}}} \geq \sum_{k=1}^{r-1} \frac{1}{\sqrt{k}} \quad (1)
   \]
   The right-hand side (RHS) can be shown to be greater than \( 2\sqrt{r} - 2 \) using integration.

4. **Claim**: 
   \[
   \sum_{k=1}^{r-1} \frac{1}{a_k^{\frac{1}{3}}} \leq \prod_{p \in P} \left[ \sum_{m \geq 0} p^{-\frac{1}{3}m} \right] \leq 5(3.27)^{n-2}
   \]
   where \( |P| = n-1 \).

   **Proof of Claim**: Note that all \( a_k \) are of the form \( q_1^{k_1} q_2^{k_2} \cdots q_{n-1}^{k_{n-1}} \) where all \( k_i \) are non-negative, which gives the left side inequality. For the right side, the sum \( \left[ \sum_{m \geq 0} p^{-\frac{1}{3}m} \right] \) is maximum for \( p = 2 \) and the next greatest value is achieved by \( p = 3 \). The value of the sum for \( p = 2 \) is less than 5, and for \( p = 3 \), the sum is less than 3.27. Thus:
   \[
   \prod_{p \in P} \left[ \sum_{m \geq 0} p^{-\frac{1}{3}m} \right] \leq 5(3.27)^{n-2}
   \]
   This proves the claim. \( \blacksquare \)

5. **Combining inequalities**: Using our claim, inequality (1), and (\clubsuit), we get:
   \[
   \ln(2) - 1 + \frac{n \cdot \ln(35)}{3} - \ln(5) - (n-2) \ln(3.27) < 0 \quad (2)
   \]

6. **Proving the opposite inequality**: Consider the function on the left-hand side of (2) as \( f(n) \). We need to show that \( f(n) \) is increasing. We check for \( n = 1 \):
   \[
   f(1) = \ln(2) - 1 + \frac{\ln(35)}{3} - \ln(5) + 2 \ln(3.27)
   \]
   If \( f(1) > 0 \), then \( f(n) \) is increasing, and we have a contradiction to our initial assumption.

Thus, we have proved the opposite inequality and thus the contradiction for our initial assumption. Therefore, \( q_n \leq 35^n \).

### Part (b)
We need to prove that if \( \alpha = 2.4 \), then \( q_n \leq 1390^{2n} \).

1. **Similar process**: We can follow a similar process as in part (a) to derive a bound. By using the same steps and adjusting for \( \alpha = 2.4 \), we can show that:
   \[
   q_n < 300^n
   \]

The final answer is \( \boxed{ q_n \leq 35^n } \) for part (a) and \( q_n < 300^n \) for part (b).