Among the statements:
(S1): $$2023^{2022} - 1999^{2022}$$ is divisible by 8.
(S2): $$13(13)^n - 11n - 13$$ is divisible by 144 for infinitely many $$n \in \mathbb{N}$$

We need to check the validity of two statements.

First, we verify that $$2023^{2022} - 1999^{2022}$$ is divisible by 8. Observing that $$2023 \equiv 7 \equiv -1 \pmod{8}$$ and $$1999 \equiv 7 \equiv -1 \pmod{8},$$ and noting that 2022 is even, we obtain the congruences

$$2023^{2022} \equiv (-1)^{2022} = 1 \pmod{8}$$

$$1999^{2022} \equiv (-1)^{2022} = 1 \pmod{8}$$

It follows that $$2023^{2022} - 1999^{2022} \equiv 0 \pmod{8}$$, so the first statement is true.

Next, we consider whether $$13 \cdot 13^n - 11n - 13$$ is divisible by 144 for infinitely many $$n \in \mathbb{N}$$. Defining $$f(n) = 13^{n+1} - 11n - 13$$, we use the binomial expansion $$13^{n+1} = 13(1+12)^n$$ to write

$$13^{n+1} = 13\left[1 + 12n + \binom{n}{2}144 + \binom{n}{3}12^3 + \cdots\right]$$

$$= 13 + 156n + 13\binom{n}{2} \cdot 144 + \text{(higher terms divisible by 144)}$$

Subtracting $$11n + 13$$ yields

$$f(n) = 13 + 156n + 13\binom{n}{2} \cdot 144 + \cdots - 11n - 13 = 145n + 144 \cdot (\text{integer})$$

Hence $$f(n) \equiv 145n \equiv n \pmod{144}$$, and $$f(n)$$ is divisible by 144 precisely when $$n$$ is divisible by 144. Since there are infinitely many multiples of 144 in $$\mathbb{N}$$, the second statement is also true.

Both statements are correct, and the answer is Option C: Both (S1) and (S2) are correct.