a) For $a = 3, b = -2, c = -1$, we have $u_{n+2} = 3u_{n+1} - 2u_n - 1, \forall n \ge 1$. By induction, one can prove
$$u_n = 2\alpha - \beta + (\beta - \alpha - 1) \cdot 2^{n-1} + n$$
for all $n \ge 1$. Then $u_{2023} = 2\alpha - \beta + (\beta - \alpha - 1)2^{2022} + 2023$.
For any $t \in \mathbb{Z}$, choose $\alpha = (2^{2022} - 1)t - 2021$ and $\beta = t - \alpha - 2$. In other words, $2 - \beta + \alpha = t$ and $\alpha + 2021 + (1 - 2^{2022})t = 0$, so we have
$$\begin{aligned} & 2\alpha - \beta + (\beta - \alpha - 1)2^{2022} + 2023 \\ &= \alpha + (2 - \beta + \alpha) + 2021 + (\beta - \alpha - 2)2^{2022} - 2^{2022} \\ &= \alpha + t + 2021 - t \cdot 2^{2022} + 2^{2022} \\ &= \alpha + 2021 + (1 - 2^{2022})t + 2^{2022} = 2^{2022}. \end{aligned}$$
This implies that there exist infinitely many pairs of $(\alpha, \beta)$ for $u_{2023} = 2^{2022}$.

b) Note that $2023 = 7 \times 17^2$. Let $(r_n)$ be the remainder of $(u_n)$ when divided by 2023. Then it is easy to check that $(r_n)$ is the periodic sequence with some period, denote by $T > 0$. We consider the following cases:

* If there exists $n_0 \in \mathbb{N}^*$ such that $7 \mid u_{n_0}$ or $17 \mid u_{n_0}$. Let consider the first case while the other case does the same. Since $7 \mid 2023$,
$$\prod_{i=0}^{m} u_{n_0 + i} \neq 1 \pmod{2023}, \forall m \ge 1.$$
Hence proposition ii) is not satisfied. For all $l \in \mathbb{N}^*$, choose $m = (2023l - 1)T$. Because $u_{n_0+nT} \equiv u_{n_0} \pmod{2023}$, and $7 \mid 2023$ so $u_{n_0+nT} \equiv u_{n_0} \equiv 0 \pmod{7}$ for all $n \in \mathbb{N}^*$.
Thus the sequence $u_{n_0}, u_{n_0+1}, \dots, u_{n_0+(2023l-1)T}$ contains at least 2023 terms which are divisible by 7. Hence,
$$\prod_{i=0}^{m} u_{n_0+i} \text{ is divisible by } 7^{2023}.$$
Therefore, proposition i) is satisfied.

* If $7 \nmid u_n, 17 \nmid u_n$ for all $n \in \mathbb{N}^*$, choose $n_0 = 1$, obviously proposition i) is not satisfied. Otherwise, $\gcd(u_n, 2023) = 1$, so by Euler's theorem,
$$u_n^{\varphi(2023)} \equiv 1 \pmod{2023}, \quad \forall n \in \mathbb{N}^*.$$
Set $a = \varphi(2023)$, for all $l \in \mathbb{N}^*$, then choose $k = laT$. We will prove that $2023 \mid \prod_{i=0}^{k} u_{1+i} - 1$. Indeed, we have
$$u_1 \equiv u_{1+T} \equiv \cdots \equiv u_{1+(la-1)T} \pmod{2023}$$
$$u_2 \equiv u_{2+T} \equiv \cdots \equiv u_{2+(la-1)T} \pmod{2023}$$
......
$$u_T \equiv u_{2T} \equiv \cdots \equiv u_{laT} \pmod{2023}.$$
Hence $u_i u_{i+T} \cdots u_{i+(la-1)T} \equiv u_i^a \equiv 1 \pmod{2023}$ for all $i = \overline{1,T}$. From this we conclude that $\prod_{i=0}^{k} u_{1+i} \equiv 1 \pmod{2023}$ or $2023 \mid \prod_{i=0}^{k} u_{1+i} - 1$. So proposition ii) is true. $\square$