The current going to/from the two parallel packs (during acceleration and charging) will not be split evenly between them, the proportion of what goes where is defined by the resistance of the packs.
The packs
will attempt to equalize when there is any difference in voltage between them. You'll obviously lose some energy to heat while this is happening, but they'll balance out to the same voltage.
Saying that putting a 4.2v 3000mah 5A cell in parallel with a 4.2v 2000mah 10A cell results in a 5A 4000mah pack just doesn't make any sense.
If we're assuming the cells have the same resistance, they'll see the same current, and the weakest cell can handle 5A, so the pack is capable of supporting at least 10A.
The 3000mah cell has 12.6Wh stored, and the 2000mah cell has 8.3Wh stored, and since they're connected in parallel they always have the same voltage. When we run both of them down it's all gone, so the 4000mah doesn't account for another 4.2Wh.
If the cells have different internal resistances, when we're drawing power from both of them in parallel we have effectively created a current divider. If the branch connnecting the 5A-capable cell (including any lengths of wire used for the connection) has twice the resistance of the 10A-capable cell, that branch is only going to experience half of the load, so you can indeed safely pull 15A out of the pack.
Current divider calculator to play around with:
http://www.learningaboutelectronics.com/Articles/Current-divider-calculator.php#answer
Because cell resistances are so low, the length of wire used can actually have a significant effect on this balance.
