Twin primes

We know $p$ and $(p+2)$ have the same parity, either both odd or both even. There is only one even prime number, that is two so if both are even, they wouldn't be primes as they are greater than three. So both primes are odd leaving $(p+1)$ to be even. Now with $p, p+2$ and $(p+1)$ in between, we have $2+1=3$ consecutive numbers. So at least one of them is divisible by three. The only prime divisible by three is three itself but both primes are greater than three so neither is divisible by three. Therefore, $(p+1)$ must be divisible by three.

Finally, $(p+1)$ is divisible by $2\times 3=6$ . Then $p≡6-1≡5$ (mod $6$ ). So $a=5, b=6$ and $a+b=\boxed{11}$