The answer is
61 + 98 = 159
We are given that there are two six sided dice and that there are 12 unique numbers on them. Some of the numbers look the same as others but they still only occur once. For example we might see 66 and 99, we don't know which is which, but we know that they are not both 66.
The problem requires us to identify which die is which in the two images for each of the three rolls.
For the sake of clarity, we will call the die on the left in roll 1 "DIE 1", and the one on the right "DIE 2". We know nothing about the sums at this point.
Notice that DIE 1 has a 69 on it. Rotated 180 degrees it is still 69 so it is unique. Therefore the die on the left in roll 3 is DIE 1. We can add the (68/89) to DIE 1. Note the orientation of the (86/98) in DIE 2 of roll 1 and DIE 2 of roll 3. It would be impossible for these two numbers to be the same based on their orientations, so the DIE 2 in roll 3 is the opposite three faces that we couldn't see in roll 1.
Now looking at roll 2 we know that the die on the right must be DIE 1 because 19/61 cannot be on DIE 2. We can now fill in all of the numbers on DIE 1.
Now we can start looking at the sums. We know that the bottom number of DIE 1 in roll 1 is 19 or 61. 105 - 61 = 44 which is not possible so the bottom number of DIE 1 is 19. This identifies the 61 in DIE 1 and tells us that the bottom of DIE 2 is 86. This also identifies the 98.
The number on the bottom of DIE 1 in roll 2 is either 18 or 81. 149 - 18 = 131 which is not possible, so the number is 81. This identifies the 18 on DIE 2 and the 68 which we already know is opposite the 86 on DIE 2.
There is only one possible way to merge the two images of DIE 2 and that is like this:
The number on the bottom of DIE 1 in roll 3 is 61 and the number on the bottom of DIE 2 in roll 3 is 98. The sum is 159.
Proof that DIE 2 can only be built one way