The polynomial we need to find is $P(x) = \boxed{-\frac{1}{5}x^2+x}$. To prove this, first we note that, if an integer $l$ is such that $4^l < i \leq 4^{l+1}$, then $$-(l+1) \leq \log_4 i < -l \Rightarrow a_i = \frac{1}{4^{-(l+1)}} = 4^{l+1}.$$Now let $n \geq 2$ be a positive integer. Let $k$ be the unique positive integer such that $4^k < n \leq 4^{k+1}$. Then \begin{align*} b_n &= \frac{1}{n^2} \left( \sum_{i=1}^{4^k} a_i +\sum_{i=4^{k}+1}^{n} a_i - \frac{1}{5} \right) \\ &= \frac{1}{n^2} \left( \left( 1+(4-1) \cdot 4 + (16-4) \cdot 4^2 + \ldots + (4^k-4^{k-1}) \cdot 4^k \right) +(n-4^k) \cdot 4^{k+1} - \frac{1}{5} \right)\\ &= \frac{1}{n^2} \left( \left( 1+3 \cdot (4 + 4^1 \cdot 4^2 + \ldots + 4^{k-1} \cdot 4^k) \right) +(n-4^k) \cdot 4^{k+1} - \frac{1}{5} \right)\\ &= \frac{1}{n^2} \left( \left( 1+3 \cdot 4 \cdot \frac{4^{2k}-1}{4^2-1} \right) +(n-4^k) \cdot 4^{k+1} - \frac{1}{5} \right) \\ &= \frac{1}{n^2} \left( \left( 1+\frac{4^{2k+1}-4}{5} \right) +(n-4^k) \cdot 4^{k+1} - \frac{1}{5} \right) \\ &= \frac{1}{n^2} \left( \frac{4^{2k+1}}{5} +n\cdot 4^{k+1} - 4^{2k+1 }\right) \\ &= \frac{1}{n^2} \left(- \frac{4^{2k+2}}{5} +n\cdot 4^{k+1}\right) \\ &= \frac{1}{n^2} \left(- \frac{a_n^2}{5} +n\cdot a_n\right) \\ &= P \left( \frac{a_n}{n} \right). \end{align*}

b) Let $\epsilon$ be a positive zero of $-\frac{1}{5}x^2+x=\frac{2024}{2025}$. Since $\frac{2024}{2025}$ lies in the intervals $(P(1), P(2)), (P(4), P(3))$, we choose the one such that $1 < \epsilon < 2$. Now, define the sequence $(n_k)$ such that $n_k = \left\lfloor \frac{4^{k+1}}{\epsilon} \right\rfloor$ for all $k \geq 1$. Since $1 < \epsilon < 2$, it follows that $4^k < n_k < 4^{k+1}$, hence $a_{n_k} = 4^{k+1}$. It is trivial to see that this sequence consists of positive integers and is increasing. Now, $\frac{4^{k+1}}{\epsilon} \leq n_k < \frac{4^{k+1}}{\epsilon} + 1$ implies $\frac{4^{k+1}}{\frac{4^{k+1}}{\epsilon} + 1} < \frac{a_{n_k}}{n_k} \leq \epsilon.$ However, note that $\lim_{k \to \infty} \frac{4^{k+1}}{\frac{4^{k+1}}{\epsilon} + 1} = \epsilon$, it follows by squeeze theorem that $\lim_{k \to \infty} \frac{a_{n_k}}{n_k} = \epsilon$, and hence from part a, $$\lim_{k \to \infty} b_{n_k} = \lim_{k \to \infty} f \left( \frac{a_{n_k}}{n_k} \right) = f(\epsilon) = \frac{2024}{2025}.$$