a) Assume to the contrary that there exists a non-zero polynomial $P(x, y) \in \mathbb{R}[x, y]$ such that for every $n \in \mathbb{N}$, $P(n, 2^n) = 0$. Let $d$ be the degree of $P$ with respect to its second variable, $y$. $P(x, y)$ can be written as,
$$P(x, y) = p_d(x)y^d + \cdots + p_1(x)y + p_0(x),$$
where $p_i$'s are polynomials in $x$ and $p_d$ is not zero.
For every $\epsilon > 0$, there exists $N_1 > 0$ such that for every $n > N_1$, $|p_d(n)| > \epsilon$. Furthermore, there exists $N_2 > 0$ such that for every $n > N_2$,
$$|p_0(n)|, \dots, |p_{d-1}(n)| < (\sqrt{2})^n = 2^{\frac{n}{2}}.$$
Now for every $n > \max\{N_1, N_2\}$,
$$\begin{align*} |P(n, 2^n)| &= |p_d(n)2^{dn} + \cdots + p_1(n)2^n + p_0(n)| \\ &\ge |p_d(n)2^{dn}| - \left( |p_{d-1}(n)2^{(d-1)n}| + \cdots + |p_1(n)2^n| + |p_0(n)| \right) \\ &> \epsilon 2^{dn} - 2^{\frac{n}{2}}(1 + 2^n + \cdots + 2^{(d-1)n}) \\ &> \epsilon 2^{dn} - 2^{(d-1)n + \frac{n}{2} + 1} \\ &= 2^{dn} (\epsilon - 2^{-\frac{n}{2}+1}). \end{align*}$$
Since $2^{-\frac{n}{2}+1}$ approaches zero as $n$ approaches infinity, $|P(n, 2^n)|$ approaches infinity and cannot be zero for large values of $n$.

b) No! Let $\alpha = \log_2 3$ (i.e., $2^\alpha = 3$). It is easy to prove that $\alpha$ is irrational. Suppose that the two mentioned sequences are Co-Algebraic and that there exists a non-zero polynomial $P(x, y)$ such that for every positive natural number $n$, $P(2^n, 3^n) = 0$. This polynomial is the sum of monomials of the form $c_{k,l}x^k y^l$, where $k$ and $l$ are nonnegative integers and $c_{k,l} \neq 0$ is a real number. Thus $P(2^n, 3^n)$ is the sum of expressions of the form $c_{k,l}2^{nk}3^{nl} = c_{k,l}2^{n(k+\alpha l)}$. Since $\alpha$ is irrational, different pairs $(k, l)$ of integers lead to different values for $k+\alpha l$. Let $\beta$ be the maximum value of $k + \alpha l$ among all of the pairs $(k, l)$ for which $c_{k,l}$ is not zero.
Therefore, $P(2^n, 3^n)$ can be written as
$$P(2^n, 3^n) = c_2 2^{\beta n} + c_1 2^{\beta_1 n} + \dots + c_k 2^{\beta_k n}, \quad (*)$$
where $c \neq 0$ and $\beta_i < \beta$ for $1 \le i \le k$. Now, dividing both sides of (*) by $2^{\beta n}$ results in
$$\frac{1}{2^{\beta n}} P(2^n, 3^n) = c + c_1 2^{(\beta_1 - \beta)n} + \dots + c_k 2^{(\beta_k - \beta)n}.$$
The left hand side approaches zero and the right hand side approaches $c$ as $n$ approaches infinity. This is a contradiction because it is assumed that $c \neq 0$, and shows that two sequences $2^n$ and $3^n$ are not Co-Algebraic.

c) According to part (b), there exists $m \in \mathbb{N}$ such that $f(2^m, 3^m) \neq 0$. Let $k$ be the largest natural number such that $5^k|P(2^m, 3^m)$. It is claimed that if $n := m + 4 \times 5^k$, then $f(2^n, 3^n)$ is not divisible by $5^{k+1}$, and since $k+1 < n$, not divisible by $5^n$ either. In order to prove it, note that $\varphi(5^{k+1}) = 4 \times 5^k$, and by Euler's theorem
$$2^{4 \times 5^k} \stackrel{5^{k+1}}{\equiv} 3^{4 \times 5^k} \stackrel{5^{k+1}}{\equiv} 1.$$
Hence
$$f(2^n, 3^n) = f(2^{m+4 \times 5^k}, 3^{m+4 \times 5^k}) \stackrel{5^{k+1}}{\equiv} f(2^m, 3^m) \stackrel{5^{k+1}}{\not\equiv} 0.$$