Find all function $f:\mathbb{Z}\to\mathbb{Z}$ satisfying

$\text{(i)}$ $f(f(m)+n)+2m=f(n)+f(3m)$ for every $m,n\in\mathbb{Z}$,

$\text{(ii)}$  there exists a $d\in\mathbb{Z}$ such that $f(d)-f(0)=2$, and

 $\text{(iii)}$ $f(1)-f(0)$ is even.
1. **Define the assertion**: Let \( P(m, n) \) be the assertion of the given functional equation:
   \[
   P(m, n): f(f(m) + n) + 2m = f(n) + f(3m)
   \]
   Let \( a = f(0) \) and \( c = f(1) \).

2. **Evaluate \( P(m, 0) \)**:
   \[
   P(m, 0): f(f(m)) + 2m = f(0) + f(3m) \implies f(f(m)) + 2m = a + f(3m)
   \]
   Rearrange to get:
   \[
   f(3m) = f(f(m)) + 2m - a
   \]

3. **New assertion \( Q(m, n) \)**:
   Using the result from step 2, we can rewrite \( P(m, n) \) as:
   \[
   f(f(m) + n) + 2m = f(n) + f(f(m)) + 2m - a
   \]
   Simplifying, we get:
   \[
   f(f(m) + n) = f(n) + f(f(m)) - a
   \]
   Let this be the new assertion \( Q(m, n) \).

4. **Evaluate \( P(m, 3m - f(m)) \)**:
   \[
   P(m, 3m - f(m)): f(f(m) + (3m - f(m))) + 2m = f(3m - f(m)) + f(3m)
   \]
   Simplifying, we get:
   \[
   f(3m) + 2m = f(3m - f(m)) + f(3m)
   \]
   Thus,
   \[
   f(3m - f(m)) = 2m
   \]
   This implies that \( 2\mathbb{Z} \subseteq f(\mathbb{Z}) \).

5. **New assertion \( R(m, n) \)**:
   Using \( Q(m, n) \), we can derive:
   \[
   f(2m + n) = f(n) + f(2m) - a
   \]
   Let this be the new assertion \( R(m, n) \).

6. **Evaluate \( R(m, 2n) \)**:
   \[
   R(m, 2n): f(2m + 2n) = f(2n) + f(2m) - a
   \]
   This shows that \( f(x) - a \) is additive over \( 2\mathbb{Z} \), implying linearity:
   \[
   f(2x) = 2kx + a \quad \text{for some integer } k
   \]

7. **Evaluate \( R(m, 1) \)**:
   \[
   R(m, 1): f(2m + 1) = f(2m) + c - a = 2km + c
   \]
   Since \( c - a \) is even, both \( f(2m) \) and \( f(2m + 1) \) are either both even or both odd.

8. **Conclusion from \( P(m, n) \)**:
   Since \( f(x) \) is even for all \( x \), both \( a \) and \( c \) are even.

9. **Evaluate \( P(2m, 0) \)**:
   \[
   P(2m, 0): f(6m) = f(f(2m)) + 4m - a
   \]
   Using \( f(2m) = 2km + a \):
   \[
   f(6m) = f(2km + a) + 4m - a
   \]
   Since \( f \) is linear:
   \[
   f(6m) = 6km + a
   \]
   Comparing coefficients, we get:
   \[
   6k = 6 \implies k = 1
   \]

10. **Final form of \( f \)**:
    \[
    f(2n) = 2n + 2u \quad \text{and} \quad f(2n + 1) = 2n + 2v \quad \forall n \in \mathbb{Z}
    \]
    where \( u, v \in \mathbb{Z} \).

The final answer is \( \boxed{f(2n) = 2n + 2u \text{ and } f(2n + 1) = 2n + 2v \quad \forall n \in \mathbb{Z}} \).