Let us say that p1, p2, p3 are distinct primes such that p1+p2=p3. Now, since p1+p2>=2+3=5, so p3>=5 and p3 is therefore an odd prime. So either p1 or p2 must be even and the other must be odd. Without loss of generality, let us say that p1 is even and p2 is odd. But then again, 2 is the only even prime, so p1=2. Taking into account that (2,3,5) is indeed a trivial solution (because 2+3=5; {2,3,5}-primes), we can deduce that other solutions are: (2,p2,p3) where p3>p2>=5.
Therefore, we can tell that neither p2 or p3 will be divisible 2 or 3 and that means that neither of them could be expressed as 6k, 6k+2, 6k+3 or 6k+4 for any integer k. That leaves us with case that if p2,p3 are indeed non-trivial primes, then they can only be expressed as 6k+1 or 6k-1 for a particular integer k (but it does not work other way around: if the integer l is in the form of 6k+1 or 6k-1, it does not necessarily mean that l is a prime).
Now, coming back to the primary equation, we actually get that: for a particular integer k: p2=6k-1 and p3=6k+1, satistying both conditions: divisibility and the equation, i.e. 2+(6k-1)=6k+1 <=> p1+p2=p3
The only question is: are there infinitely many such integers k, that it would be satisfied? Because even though we proved that non-trivial primes are in that particular form (6k-1 or 6k+1 for some integer k) but we cannot state that there are infinitely many pairs that with particular integer k that BOTH 6k-1 and 6k+1 would be primes. If someone asked me, I would guess that it would probably have infinitely many such solutions (2,6k-1,6k+1) for different integers k but how to prove it exactly?
Pegarding your question about the primes of the form $6k\pm 1$ a simple way to test it is to work in $\mathbb{Z}_6$ and see for which cases we have odd numbers, can you see this post.
Now, regarding your question about twin primes, as you have already been told, it is an open problem and it seems that we are a long way from proving it. As they told you in the comments, the minimum gap is currently $246$ (result of the Polymath project) and assuming the Elliott-Halberstan conjecture the gap can be reduced to $16$. Of course, the optimal result is to be able to prove that $\lim\inf_{n\to\infty} (p_{n+1}-p_n)=2$.