Find all functions $f:\mathbb{Z} \to \mathbb{Z}$ such that for all $m\in\mathbb{Z}$: [list][*] $f(m+8) \le f(m)+8$, [*] $f(m+11) \ge f(m)+11$.[/list]
1. We start with the given conditions for the function \( f: \mathbb{Z} \to \mathbb{Z} \):
   \[
   f(m+8) \leq f(m) + 8
   \]
   \[
   f(m+11) \geq f(m) + 11
   \]

2. From the first condition, we have:
   \[
   f(m+8) \leq f(m) + 8
   \]
   Adding 3 to both sides, we get:
   \[
   f(m+8) + 3 \leq f(m) + 11
   \]
   Which can be rewritten as:
   \[
   f(m) + 11 \geq f(m+8) + 3
   \]

3. From the second condition, we have:
   \[
   f(m+11) \geq f(m) + 11
   \]
   Combining this with the result from step 2, we get:
   \[
   f(m+11) \geq f(m+8) + 3
   \]

4. This implies:
   \[
   f(m) \geq f(m-3) + 3 \quad (1)
   \]

5. Now, take a fixed integer \( m_0 \). Using inequality (1) and the first condition, we derive the following:
   \[
   f(m_0 + 2) + 6 \leq f(m_0 + 8) \leq f(m_0) + 8
   \]
   \[
   f(m_0) + 9 \leq f(m_0 + 9) \leq f(m_0 + 1) + 8
   \]
   \[
   f(m_0 + 1) + 9 \leq f(m_0 + 10) \leq f(m_0 + 2) + 8
   \]

6. Summarizing the inequalities:
   \[
   f(m_0 + 2) + 6 \leq f(m_0) + 8
   \]
   \[
   f(m_0) + 9 \leq f(m_0 + 1) + 8
   \]
   \[
   f(m_0 + 1) + 9 \leq f(m_0 + 2) + 8
   \]

7. Adding the left-hand sides and the right-hand sides of these inequalities, we get:
   \[
   (f(m_0 + 2) + 6) + (f(m_0) + 9) + (f(m_0 + 1) + 9) \leq (f(m_0) + 8) + (f(m_0 + 1) + 8) + (f(m_0 + 2) + 8)
   \]
   Simplifying, we find:
   \[
   f(m_0 + 2) + f(m_0) + f(m_0 + 1) + 24 \leq f(m_0) + f(m_0 + 1) + f(m_0 + 2) + 24
   \]
   This implies:
   \[
   f(m_0 + 2) + f(m_0) + f(m_0 + 1) + 24 = f(m_0) + f(m_0 + 1) + f(m_0 + 2) + 24
   \]
   Therefore, the inequalities must be equalities:
   \[
   f(m_0 + 2) + 6 = f(m_0) + 8
   \]
   \[
   f(m_0) + 9 = f(m_0 + 1) + 8
   \]
   \[
   f(m_0 + 1) + 9 = f(m_0 + 2) + 8
   \]

8. Solving these equalities, we get:
   \[
   f(m_0 + 2) = f(m_0) + 2
   \]
   \[
   f(m_0) = f(m_0 + 1) - 1
   \]
   \[
   f(m_0 + 1) = f(m_0 + 2) - 1
   \]

9. This implies:
   \[
   f(m_0) = f(m_0 + 1) - 1 = f(m_0 + 2) - 2
   \]
   Which holds for any integer \( m \). Hence, it concludes that:
   \[
   f(m+1) = f(m) + 1
   \]
   for all integers \( m \).

10. Therefore, the function \( f \) must be of the form:
    \[
    f(m) = m + a
    \]
    for some integer \( a \).

11. Finally, we verify that this form satisfies the original conditions:
    \[
    f(m+8) = (m+8) + a = f(m) + 8
    \]
    \[
    f(m+11) = (m+11) + a = f(m) + 11
    \]
    Both conditions are satisfied.

The final answer is \( \boxed{ f(m) = m + a } \) for some integer \( a \).