Finding multiple smaller collisions equal work as finding a bigger collision?
Are problem A and B equivalent?  In other words, do both take $2^{256}$ work?

No; in the first case, you can find the first collision with $2^{128}$ effort and then find the second collision with $2^{128}$ effort, yielding a total effort of $2^{128}+2^{128} = 2^{129}$
Actually, you'd likely start hashing arbitrary values, and stop after both a second preimage of $x$ and a second preimage of $y$ came up; this expected effort is slightly less than $2^{129}$...