Consider the sequence $(a_n)$ satisfying $a_1=\dfrac{1}{2},a_{n+1}=\sqrt[3]{3a_{n+1}-a_n}$ and $0\le a_n\le 1,\forall n\ge 1.$

a. Prove that the sequence $(a_n)$ is determined uniquely and has finite limit.

b. Let $b_n=(1+2.a_1)(1+2^2a_2)...(1+2^na_n), \forall n\ge 1.$

Prove that the sequence $(b_n)$ has finite limit.
### Part (a): Prove that the sequence $(a_n)$ is determined uniquely and has a finite limit.

1. **Define the sequence and the recursive relation:**
   Given the sequence $(a_n)$ with $a_1 = \frac{1}{2}$ and the recursive relation:
   \[
   a_{n+1} = \sqrt[3]{3a_{n+1} - a_n}
   \]
   We need to show that this sequence is uniquely determined and has a finite limit.

2. **Reformulate the recursive relation:**
   Consider the equation:
   \[
   x^3 - 3x + a_n = 0
   \]
   We need to find $a_{n+1}$ such that $a_{n+1} = x$ and $x \in [0, 1]$.

3. **Analyze the cubic equation:**
   The function $f(x) = x^3 - 3x + a_n$ is strictly decreasing on $[0, 1]$ because:
   \[
   f'(x) = 3x^2 - 3
   \]
   For $x \in [0, 1]$, $f'(x) \le 0$. Therefore, $f(x)$ is strictly decreasing on $[0, 1]$.

4. **Existence and uniqueness of the root:**
   Since $f(x)$ is strictly decreasing and continuous on $[0, 1]$, and $f(0) = a_n \ge 0$ and $f(1) = 1 - 3 + a_n = a_n - 2 \le -1$, by the Intermediate Value Theorem, there exists a unique $x \in [0, 1]$ such that $f(x) = 0$. Hence, $a_{n+1}$ is uniquely determined.

5. **Monotonicity and boundedness:**
   Since $a_{n+1}$ is the unique root of $x^3 - 3x + a_n = 0$ in $[0, 1]$, and $f(x)$ is strictly decreasing, we have:
   \[
   a_{n+1} \le a_n
   \]
   Therefore, the sequence $(a_n)$ is non-increasing and bounded below by 0.

6. **Convergence:**
   Since $(a_n)$ is non-increasing and bounded below, it converges to some limit $l \in [0, 1]$. Taking the limit as $n \to \infty$ in the recursive relation:
   \[
   l = \sqrt[3]{3l - l} \implies l = \sqrt[3]{2l}
   \]
   Cubing both sides:
   \[
   l^3 = 2l \implies l(l^2 - 2) = 0
   \]
   Since $0 \le l \le 1$, the only solution is $l = 0$.

   Therefore, the sequence $(a_n)$ converges to 0.

   \[
   \boxed{\lim_{n \to +\infty} a_n = 0}
   \]

### Part (b): Prove that the sequence $(b_n)$ has a finite limit.

1. **Define the sequence $(b_n)$:**
   Given $b_n = (1 + 2 \cdot a_1)(1 + 2^2 \cdot a_2) \cdots (1 + 2^n \cdot a_n)$ for all $n \ge 1$.

2. **Behavior of $a_n$:**
   From part (a), we know that $a_n \to 0$ as $n \to \infty$.

3. **Examine the terms of $b_n$:**
   Each term in the product defining $b_n$ is of the form $1 + 2^k \cdot a_k$. Since $a_k \to 0$, for large $k$, $2^k \cdot a_k \to 0$.

4. **Convergence of the product:**
   For large $n$, $1 + 2^k \cdot a_k \approx 1$. Therefore, the product of these terms will converge to a finite limit because the terms $2^k \cdot a_k$ become negligible as $k$ increases.

   Hence, the sequence $(b_n)$ has a finite limit.

   \[
   \boxed{\lim_{n \to +\infty} b_n \text{ is finite}}
   \]