Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function which satisfies the following:
[list][*] $f(m)=m$, for all $m\in\mathbb{Z}$;[*] $f(\frac{a+b}{c+d})=\frac{f(\frac{a}{c})+f(\frac{b}{d})}{2}$, for all $a, b, c, d\in\mathbb{Z}$ such that $|ad-bc|=1$, $c>0$ and $d>0$;[*] $f$ is monotonically increasing.[/list]
(a) Prove that the function $f$ is unique.
(b) Find $f(\frac{\sqrt{5}-1}{2})$.
1. **Claim:** If rationals \( r, s \) are adjacent, then for any non-negative integer \( n \): 
   \[
   f((n.r)*s) = \frac{1}{2^n} \left( (2^n - 1) f(r) + f(s) \right)
   \]
   **Proof:** Let \( r = \frac{a}{b} \) and \( s = \frac{c}{d} \) (with \( b, d > 0 \)). We proceed by induction on \( n \).

   - **Base Case:** For \( n = 0 \), we have:
     \[
     f((0.r)*s) = f(s) = \frac{1}{2^0} \left( (2^0 - 1) f(r) + f(s) \right) = f(s)
     \]
     which is trivially true.

   - **Inductive Step:** Assume the statement is true for some \( n \). We need to show it holds for \( n + 1 \). Observe that \( (n.r)*s \) and \( r \) are adjacent because:
     \[
     |(na + c)b - (nb + d)a| = |bc - ad| = 1
     \]
     So we can apply the second condition to get:
     \[
     f(((n+1).r)*s) = \frac{1}{2} \left( f((n.r)*s) + f(r) \right)
     \]
     By the inductive hypothesis:
     \[
     f((n.r)*s) = \frac{1}{2^n} \left( (2^n - 1) f(r) + f(s) \right)
     \]
     Therefore:
     \[
     f(((n+1).r)*s) = \frac{1}{2} \left( \frac{1}{2^n} \left( (2^n - 1) f(r) + f(s) \right) + f(r) \right)
     \]
     Simplifying:
     \[
     f(((n+1).r)*s) = \frac{1}{2^{n+1}} \left( (2^n - 1) f(r) + f(s) + 2^n f(r) \right) = \frac{1}{2^{n+1}} \left( (2^{n+1} - 1) f(r) + f(s) \right)
     \]
     This completes the induction.

2. Given a real number \( x \), consider its continued fraction \( x = [a_0, a_1, a_2, \ldots] \) (where \( a_0 = \lfloor x \rfloor \) and \( a_k \in \mathbb{N} \) for all \( k > 0 \)). Define the convergents \( c_n = [a_0, \ldots, a_n] \) for all \( n \geq 0 \).

3. Define \( A_k = \sum_{i=1}^{k} a_i \) for all \( k \geq 1 \). If we set \( (p_0, q_0) = (a_0, 1) \), \( (p_1, q_1) = (a_0 a_1 + 1, a_1) \), and \( (p_n, q_n) = (a_n p_{n-1} + p_{n-2}, a_n q_{n-1} + q_{n-2}) \) for all \( n \geq 2 \), then:
   - \( c_n = \frac{p_n}{q_n} \) for all \( n \geq 0 \).
   - \( p_n q_{n-1} - q_n p_{n-1} = (-1)^{n-1} \) (so \( c_n \) and \( c_{n-1} \) are adjacent) for all \( n > 0 \).
   - \( p_n q_{n-2} - p_{n-2} q_n = (-1)^n a_n \) for all \( n > 1 \).

4. We have \( f(c_0) = a_0 \), \( f(c_1) = f\left( ((a_1-1).c_0) * (a_0+1) \right) = a_0 + \frac{1}{2^{a_1-1}} \). Using \( c_n = (a_n.c_{n-1})*c_{n-2} \), we can prove by induction:
   \[
   f(c_n) = a_0 - 2 \sum_{k=1}^{n} \frac{(-1)^k}{2^{A_k}}
   \]
   for all \( n > 0 \).

5. **Case 1:** \( x \) is rational. Then \( x = [a_0, \ldots, a_m] \) for some \( m \geq 0 \). If \( m = 0 \), then \( x \) is an integer and \( f(x) = x \). Otherwise:
   \[
   f(x) = f(c_m) = a_0 - 2 \sum_{k=1}^{m} \frac{(-1)^k}{2^{A_k}}
   \]
   which is uniquely determined.

6. **Case 2:** \( x \) is irrational. Then the continued fraction is infinite. We have \( x = \lim_{n \to \infty} c_n \). From the properties of continued fractions, we get \( (c_{2n}) \) (resp. \( (c_{2n+1}) \)) is increasing (resp. decreasing), so \( c_{2n} \leq x \leq c_{2n+1} \) for all \( n \geq 0 \). Since \( f \) is increasing, it follows that:
   \[
   f(c_{2n}) \leq f(x) \leq f(c_{2n+1})
   \]
   for all \( n \geq 0 \).

7. But \( f(c_n) = a_0 - 2 \sum_{k=1}^{n} \frac{(-1)^k}{2^{A_k}} \) for all \( n \geq 1 \), and \( A_k \geq k \), so \( \sum_{k=1}^{n} \frac{(-1)^k}{2^{A_k}} \) is convergent. Therefore, \( (f(c_n)) \) converges to:
   \[
   a_0 - 2 \sum_{n \geq 1} \frac{(-1)^n}{2^{A_n}}
   \]
   We conclude that:
   \[
   f(x) = a_0 - 2 \sum_{n \geq 1} \frac{(-1)^n}{2^{A_n}}
   \]
   which is uniquely determined.

The final answer is \( \boxed{ f(x) = a_0 - 2 \sum_{n \geq 1} \frac{(-1)^n}{2^{A_n}} } \)