a. With $x, y \in \mathbb{R}$, $P(x, y)$ indicates the proposition containing the variable as follows
$$f(x + g(y)) = x f(y) + (2023 - y) f(x) + g(x).$$
From $P(0, y)$, we deduce
$$f(g(y)) = (2023 - y) f(0) + g(0) = 2022(2023 - y) + g(0), \quad \forall y \in \mathbb{R}.$$
Note that the right hand side of the above equality is a polynomial of first degree in the variable $y$, so it takes all values on the set of real numbers, in other words $f$ is surjective.

We show that $g$ is injective. Consider $x_1, x_2 \in \mathbb{R}$ such that $g(x_1) = g(x_2)$, then from $P(0, x_1)$ and $P(0, x_2)$, we get
$$2022(2023 - x_1) + g(0) = 2022(2023 - x_2) + g(0) \text{ or } x_1 = x_2.$$
Thus $g$ is injective, the proof is complete.

b. Now consider the following equation
$$f(g(y)) = (2023 - y) f(0) + g(0) = 2022(2023 - y) + g(0). \quad (1)$$
From $P(g(x), y)$, we deduce
$$f(g(x) + g(y)) = g(x) f(y) + (2023 - y) f(g(x)) + g(g(x)), \quad \forall x, y \in \mathbb{R}.$$
Combined with (1), one can get $f(g(x) + g(y)) = g(x) f(y) + (2023 - y)[2022(2023 - x) + g(0)] + g(g(x))$. Swap $x, y$ in the above equation and then compare, we get
$$\begin{aligned} & g(x) f(y) + g(g(x)) + (2023 - y) g(0) \\ = & g(y) f(x) + g(g(y)) + (2023 - x) g(0), \quad \forall x, y \in \mathbb{R}. \end{aligned} \quad (2)$$
Since $f$ is surjective, there exists a real number $a$ such that $f(a) = 0$. Substituting $y = a$ in (2), we get
$$g(g(x)) = -g(0) x + g(a) f(x) + C, \quad \forall x \in \mathbb{R}$$
where $C$ is a constant. Substitute back to (2), get $g(x) f(y) - g(0) x + g(a) f(x) + C + (2023 - y) g(0) = g(y) f(x) - g(0) y + g(a) f(y) + C + (2023 - x) g(0)$, for all $x, y \in \mathbb{R}$. Simplifying this to get
$$g(x) f(y) + g(a) f(x) = g(y) f(x) + g(a) f(y), \quad \forall x, y \in \mathbb{R}.$$
Substitute $y = 0$ into the above equation
$$2022 g(x) + g(a) f(x) = g(0) f(x) + 2022 g(a)$$
thus
$$g(x) = \frac{g(0) - g(a)}{2022} f(x) + g(a), \quad \forall x \in \mathbb{R}.$$
If $g(a) = g(0)$ then $a = 0$ or $f(a) = f(0) = 0$, contradicts $f(0) = 2022$. So $g(a) \neq g(0)$, from which it follows that $g$ is surjective. Then there exists some value $b$ such that $g(b) = 0$. From $P(x, b)$ we get
$$f(x) = x f(b) + (2023 - b) f(x) + g(x), \quad \forall x \in \mathbb{R}.$$
Substituting back to $P(x, y)$ we get
$$\begin{aligned} f(x + g(y)) &= x f(y) + (2023 - y) f(x) + f(x) \\ &\quad - x f(b) + (b - 2023) f(x) \\ &= x(f(y) - f(b)) + f(x)(1 + b - y), \quad \forall x, y \in \mathbb{R}. \end{aligned}$$
Substituting $y = 1 + b$ to get
$$f(x + g(1 + b)) = x(f(1 + b) - f(b)), \quad \forall x \in \mathbb{R}$$
or $f$ is linear over $\mathbb{R}$, and the same for $g$. At this point, we just need to replace it with $P(x, y)$ to complete the solution. There are two pairs of $(f(x), g(x))$ as follows
$$\begin{cases} f(x) = \frac{-1 \pm \sqrt{5}}{2} x + 2022, \\ g(x) = 1011(-1 \mp \sqrt{5}) x + 2 \cdot 1011^2 \cdot (-3 \mp \sqrt{5}) \end{cases}, \quad \forall x \in \mathbb{R}.$$
$\Box$