Here's my attempt on the first part of the question, I'm not sure how valid it is, though; so I'd appreciate feedback/corrections:
It is known that: $\pi(n)\sim\frac{n}{\ln{n}}$. Where $\pi(n)$ is the prime counting function on $n$ (i.e. the number of primes less than or equal to $n$).
Therefore, we are looking to show that there exists some $n$ such that:
$\frac{(n+1)^{2}}{2\ln{(n+1)}}-\frac{n^{2}}{2\ln{n}}\geqslant1000$
Let us call $\pi_{d}(n)=\frac{(n+1)^{2}}{2\ln{(n+1)}}-\frac{n^{2}}{2\ln{n}}$. Therefore, we have:
$\frac{d\pi_{d}}{dn}=\left(\frac{n}{2\left(\ln{n}\right)^{2}}+\frac{n+1}{\ln{(n+1)}}\right)-\left(\frac{n}{\ln{n}}+\frac{n+1}{2\left(\ln{(n+1)}\right)^{2}}\right)$
Therefore $\frac{d\pi_{d}}{dn}\gt0$, $\forall n\gt0$. Therefore, as $\pi_{d}(n)$ is monotonically increasing: $\exists n:\pi((n+1)^{2})-\pi(n^{2})\geqslant1000$.