Find all functions $f: (0, +\infty)\cap\mathbb{Q}\to (0, +\infty)\cap\mathbb{Q}$ satisfying thefollowing conditions:
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[*] $f(ax) \leq (f(x))^a$, for every $x\in (0, +\infty)\cap\mathbb{Q}$ and $a \in (0, 1)\cap\mathbb{Q}$
[*] $f(x+y) \leq f(x)f(y)$, for every $x,y\in (0, +\infty)\cap\mathbb{Q}$
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1. **Define the function \( g(x) \):**
   Let \( g(x) = \ln(f(x)) \). Then \( f(x) = e^{g(x)} \).

2. **Transform Condition 1:**
   Condition 1 states \( f(ax) \leq (f(x))^a \) for every \( x \in (0, +\infty) \cap \mathbb{Q} \) and \( a \in (0, 1) \cap \mathbb{Q} \).
   \[
   f(ax) \leq (f(x))^a \implies e^{g(ax)} \leq e^{ag(x)} \implies g(ax) \leq ag(x)
   \]

3. **Transform Condition 2:**
   Condition 2 states \( f(x+y) \leq f(x)f(y) \) for every \( x, y \in (0, +\infty) \cap \mathbb{Q} \).
   \[
   f(x+y) \leq f(x)f(y) \implies e^{g(x+y)} \leq e^{g(x)}e^{g(y)} \implies g(x+y) \leq g(x) + g(y)
   \]

4. **Combine the inequalities:**
   We have \( g(ax) \leq ag(x) \) and \( g(x+y) \leq g(x) + g(y) \). To find \( g(x) \), consider the function \( g \) on rational numbers.

5. **Use Jensen's Inequality:**
   For \( x, y \in (0, +\infty) \cap \mathbb{Q} \), we can write:
   \[
   g(x) + g(y) = g\left(\frac{x}{x+y} \cdot (x+y)\right) + g\left(\frac{y}{x+y} \cdot (x+y)\right) \leq \frac{x}{x+y}g(x+y) + \frac{y}{x+y}g(x+y) = g(x+y)
   \]
   Since \( g(x+y) \leq g(x) + g(y) \), we have:
   \[
   g(x) + g(y) = g(x+y)
   \]

6. **Characterize \( g(x) \):**
   The functional equation \( g(x) + g(y) = g(x+y) \) is characteristic of a linear function. Therefore, \( g(x) = kx \) for some constant \( k \).

7. **Determine \( k \):**
   Since \( g(ax) \leq ag(x) \), substituting \( g(x) = kx \) gives:
   \[
   g(ax) = k(ax) = akx \leq ag(x) = akx
   \]
   This inequality holds for any \( k \). Therefore, \( k \) can be any real number.

8. **Express \( f(x) \):**
   Since \( g(x) = kx \), we have:
   \[
   f(x) = e^{g(x)} = e^{kx}
   \]

9. **Check the conditions for \( f(x) = e^{kx} \):**
   - For \( f(ax) \leq (f(x))^a \):
     \[
     f(ax) = e^{k(ax)} = (e^{kx})^a = (f(x))^a
     \]
     This holds as an equality.
   - For \( f(x+y) \leq f(x)f(y) \):
     \[
     f(x+y) = e^{k(x+y)} = e^{kx}e^{ky} = f(x)f(y)
     \]
     This also holds as an equality.

10. **Special case for \( k = 0 \):**
    If \( k = 0 \), then \( f(x) = e^{0 \cdot x} = 1 \).

11. **Rationality condition:**
    Since \( f \) maps positive rationals to positive rationals, \( f(x) = e^{kx} \) must be rational for all rational \( x \). The only way \( e^{kx} \) is rational for all rational \( x \) is if \( k = 0 \), making \( f(x) = 1 \).

Conclusion:
The only function that satisfies the given conditions is \( f(x) = 1 \).

The final answer is \( \boxed{ f(x) = 1 } \).