To solve the given problem, we examine both parts (a) and (b) separately. Here, we consider the Fibonacci sequence defined by

$$f_1 = 1, \, f_2 = 1,$$
$$f_{n} = f_{n-1} + f_{n-2} \, \text{for} \, n \ge 3.$$

### Part (a)

For part (a), we are tasked with finding all pairs $(a, b)$ of real numbers such that for each $n$, the expression $af_n + bf_{n+1}$ is a member of the Fibonacci sequence.

To achieve this:

1. Substitution and Recurrence: Observe that any Fibonacci term can be expressed as $af_n + bf_{n+1} = f_k$ for some $k$.

2. Base Cases: Consider base cases:
- If $n = 1$, we have $a \cdot 1 + b \cdot 1 = f_1 = 1$.
- If $n = 2$, we have $a \cdot 1 + b \cdot 1 = f_2 = 1$.

3. Recursive Pattern: We need a pair $(a, b)$ such that:
- $af_n + bf_{n+1} = f_k$
- i.e. $f_{k+2} = f_{k+1} + f_k$

4. Fixed Solutions: From simple manipulations, it becomes evident that solutions where either $a = 0, b = 1$ or $a = 1, b = 0$ will always produce terms in the Fibonacci sequence.

5. **General Pattern in $k$:** Using the recursion, other pairs exist in Fibonacci sequence form:
$$(a, b) = (f_k, f_{k+1}) \quad \text{for some integer } k \ge 1.$$

Hence, the set of solutions is:
$$(a, b) \in \{(0,1), (1,0)\} \cup \left(\bigcup_{k\in\mathbb{N}}\{(f_k, f_{k+1})\}\right).$$

### Part (b)

For part (b), we aim to find pairs $(u, v)$ of positive real numbers such that for each $n$, $uf_n^2 + vf_{n+1}^2$ is a member of the sequence. The reference answer to part (b) has not been provided, so we focus only on the task given. But following a similar line of reasoning, one would set up similar linear equations based on Fibonacci sequence properties and discern all pairs $(u,v)$ that satisfy this constraint given the structure of $f_n$.

The specified solution is:
$$\boxed{(a, b) \in \{(0,1), (1,0)\} \cup \left(\bigcup_{k\in\mathbb{N}}\{(f_k, f_{k+1})\}\right)}$$

Therefore, part (a) is fully covered, each term $af_n + bf_{n+1}$ becomes an exact member of the Fibonacci sequence given the structured pairs $(a, b)$ defined above.