[sp] Write it as $\Bigl(1-\dfrac1{a+1}\Bigr) + \Bigl(1+\dfrac1b\Bigr) = 2 - \dfrac4{k+2}$, which implies $\dfrac1{a+1} - \dfrac1b = \dfrac4{k+2}$. The right side of that last equation is positive, hence so is the left side. Therefore $b>a+1$.
Now multiply out the fractions, to get $(b-a-1)(k+2) = 4b(a+1)$. After playing around with that equation for a while, I decided that the best way to re-write it is as $$\bigl(b - (a+1)\bigr)\bigl(k - (4a+2)\bigr) = 4(a+1)^2.\qquad(*)$$ There are then two cases to consider.
Case 1: $a=2$. Then (*) becomes $(b-3)(k-10) = 36$. The only primes $b$ for which $b-3$ is a factor of $36$ are $5$ and $7$. These lead to the solutions $(a,b,k) = (2,5,28)$ and $(a,b,k) = (2,7, 19)$, the values of $b-a$ being $3$ and $5$.
Case 2: $a$ is an odd prime. Then $b - (a+1)$ is also odd. By (*), it is a divisor of $4(a+1)^2$, but it has no divisors in common with $4$ (apart from $1$). It also has no divisors in common with $a+1$ (apart from $1$), because any such divisor would be a nontrivial factor of $b$, contradicting the fact that $b$ is prime. The only remaining possibility is that $b - (a+1) = 1$, and then $b-a = 2$. There are (infinitely?) many such solutions, because $(a,b)$ can be any prime pair $(p,p+2)$.
In conclusion, the possible values of $b-a$ are $2,3,5$.[/sp]
