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Question 1. $f_3(x)$ is not a polynomial function, which can be seen as follows. Call $b_n$ a champion if it exceeds all the earlier terms $b_1,\dots,b_{n-1}$. Clearly, there are infinitely many champions, and they tend to infinity. Let $m$ be fixed, and consider the champions $b_n$ with $n>m$. We get, as $b_n$ tends to infinity, $$ f_3(b_n)>\prod_{k=1}^m (b_n-b_k)^3 = (1+o(1))\ b_n^{3m}. $$ This shows that $f_3$ has degree at least $3m$. As $m$ is arbitrary, the claim follows.

Question 2. You cannot show that $g(x)$ is not a polynomial function, even under the stronger condition that each $a_k(y)=a_k$ is a constant. Indeed, one can arrange that $g(x)$ is the identity map on $\mathbb{Q}$ under this condition, e.g. take $a_0:=b_1$, $a_1:=1$, $a_n:=0$ for $n\geq 2$, so that $$ G(x,y) = b_1 + (x-b_1) = x.$$ I used Julian Rosen's simplification of my original solution here.

$f_3(x)$ is not a polynomial function, which can be seen as follows. Call $b_n$ a champion if it exceeds all the earlier terms $b_1,\dots,b_{n-1}$. Clearly, there are infinitely many champions, and they tend to infinity. Let $m$ be fixed, and consider the champions $b_n$ with $n>m$. We get, as $b_n$ tends to infinity, $$ f_3(b_n)>\prod_{k=1}^m (b_n-b_k)^3 = (1+o(1))\ b_n^{3m}. $$ This shows that $f_3$ has degree at least $3m$. As $m$ is arbitrary, the claim follows.

You cannot show that $g(x)$ is not a polynomial, even under the stronger conditions that $\lambda=1$ and each $a_k(y)=a_k$ is a constant. Indeed, one can arrange that $g(x)$ is the identity map on $\mathbb{Q}$ under these conditions, i.e. that we have $$b_1=a_0,\quad b_2=a_0+a_1(b_2-b_1),\quad b_3=a_0+a_1(b_3-b_1)+a_2(b_3-b_1)(b_2-b_3),$$ and so on. This gives a recursion for $a_0,a_1,a_2,\dots$ in terms of $b_1,b_2,b_3,\dots$, and the claim follows.

Added. As Julian Rosen nicely observed, the above construction actually gives $a_0=b_1$, $a_1=1$, $a_n=0$ for $n\geq 2$, so that $$ G(x,y) = b_1 + (x-b_1) = x$$ is a finite sum.

Question 1. $f_3(x)$ is not a polynomial function, which can be seen as follows. Call $b_n$ a champion if it exceeds all the earlier terms $b_1,\dots,b_{n-1}$. Clearly, there are infinitely many champions, and they tend to infinity. Let $m$ be fixed, and consider the champions $b_n$ with $n>m$. We get, as $b_n$ tends to infinity, $$ f_3(b_n)>\prod_{k=1}^m (b_n-b_k)^3 = (1+o(1))\ b_n^{3m}. $$ This shows that $f_3$ has degree at least $3m$. As $m$ is arbitrary, the claim follows.

Question 2. You cannot show that $g(x)$ is not a polynomial function, even under the stronger condition that each $a_k(y)=a_k$ is a constant. Indeed, one can arrange that $g(x)$ is the identity map on $\mathbb{Q}$ under this condition, e.g. take $a_0:=b_1$, $a_1:=1$, $a_n:=0$ for $n\geq 2$, so that $$ G(x,y) = b_1 + (x-b_1) = x.$$ I used Julian Rosen's simplification of my original solution here.

You cannot show that $g(x)$ is not a polynomial, even under the stronger conditions that $\lambda=1$ and each $a_k(y)=a_k$ is a constant. Indeed, one can arrange that $g(x)$ is the identity map on $\mathbb{Q}$ under these conditions, i.e. that we have $$b_1=a_0,\quad b_2=a_0+a_1(b_2-b_1),\quad b_3=a_0+a_1(b_3-b_1)+a_2(b_3-b_1)(b_2-b_3),$$ and so on. This gives a recursion for $a_0,a_1,a_2,\dots$ in terms of $b_1,b_2,b_3,\dots$, and the claim follows.

You cannot show that $g(x)$ is not a polynomial, even under the stronger conditions that $\lambda=1$ and each $a_k(y)=a_k$ is a constant. Indeed, one can arrange that $g(x)$ is the identity map on $\mathbb{Q}$ under these conditions, i.e. that we have $$b_1=a_0,\quad b_2=a_0+a_1(b_2-b_1),\quad b_3=a_0+a_1(b_3-b_1)+a_2(b_3-b_1)(b_2-b_3),$$ and so on. This gives a recursion for $a_0,a_1,a_2,\dots$ in terms of $b_1,b_2,b_3,\dots$, and the claim follows.

Added. As Julian Rosen nicely observed, the above construction actually gives $a_0=b_1$, $a_1=1$, $a_n=0$ for $n\geq 2$, so that $$ G(x,y) = b_1 + (x-b_1) = x$$ is a finite sum.