Forgive my very naive question. I know next to nothing about number theory, but I'm curious about the state of the art on the distribution of primes.

Let $\mathrm{Li}(x)$ be the offset logarithmic integral, let $\pi(x)$ be the prime counting function, and recall that $$ \pi(x) \sim \mathrm{Li}(x). $$ For any $a,n\in\mathbb{N}$, let $$ I_n(a) \;=\; \bigl[a ,\ \mathrm{Li}^{-1}\bigl(\mathrm{Li}(a)+n\bigr)\bigr] $$ That is, $I_n(a)$ is the interval starting at $a$ in which we expect to find $n$ primes. Let $N_n(a)$ denote the actual number of primes in the interval $I_n(a)$, which defines a function $N_n\colon \mathbb{N}\to\mathbb{N}$.

Now, consider the following statement.

For all $n,k\in\mathbb{N}$, $$ \lim_{M\to\infty} \frac{\bigl|N_n^{-1}(\{k\}) \cap \{1,\ldots,M\}\bigr|}{M} \;=\; \frac{n^k e^{-n}}{k!}.\tag{1} $$

The quantity on the left is roughy the probability that, instead of finding $n$ primes in the interval $I_n(a)$, we find exactly $k$. The formula on the right is a Poisson distribution with parameter $n$. It is the naive thing one would expect this limit to converge to, assuming each number $m$ is prime with probability $1/\log(m)$.

Applying something like the central limit theorem to the above approximation, one obtains the statement

For all $x\in\mathbb{R}$, $$ \lim_{n\to\infty} \lim_{M\to\infty} \frac{\bigl|N_n^{-1}([0,n+x\sqrt{n}]) \cap \{1,\ldots,M\}\bigr|}{M} \;=\; \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-t^2/2}dt\tag{2} $$

where the formula on the right is the cumulative distribution function of a standard normal distribution.

My question is:

What is the status of statements (1) and (2)?

In particular, are they (a) known to be true, (b) open but weaker than the Riemann hypothesis, (c) equivalent to the Riemann hypothesis, (d) stronger than the Riemann hypothesis, (e) unrelated to the Riemann hypothesis, or (f) known to be false?