First puzzle:

By the link with the 4, the blues must be either threes or fives. If they are threes, then the reds are 10 and 11, and the up-triangle containing both reds has a sum that is too large. Therefore the blues are 5. The first green circle is then a 3, because both fives are taken. For the other two, their sum must be 7, so they can only be 3 and 4. Filling in the rest of the yellow circles, we can see that the rightmost circle in the bottom row must be an 8, for if it were a 6 we would need to use an extra 4.

Second puzzle:
Blues are obviously 7. Now the leftmost yellow and the red circles must be 6 and 6, or 8 and 4. If they are both sixes, then the fact that the yellows are the same as the oranges leads to a contradiction, as we can't use any more sixes. So red is 4, and yellows must be 8 and 9 (because 7 is already taken), oranges also being 8 and 9. This allows us to fill all green squares. Then the topmost circle is either 6 or 4; but if it is 6 then we need an extra 8 which we can't use, so it must be 4. The rest can be then filled out:
