Personally, I would have solved it as follows.
First, calculated the inital velocity during the2nd second of falling:
d = (Vi)t + .5at^2
5 = (Vi)*1 + 0.5a(1)(1)
Vi = 5 - 0.5a
Since the inital velocity in the 2nd second of falling is the same as the final velocity in the 1st second of falling, you can subsitute Vi above for Vf in the equation below:
a= (Vf - Vi) / t
a = (5 - 0.5a - 0) / 1
a = 5 - 0.5a
1.5a = 5
a = 3.333
Just offering an alternate solution....
Edited for clarity
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