First Thoughts:

At least one of the dice must have a 0 on it. Since we are still free to choose, let’s choose die number 1. We observe that since die number 2 can only have six digits on it, no matter which six digits we choose, we cannot obtain all of the dates with a 0 in them. For example, if we chose die number 2 to have (1, 2, 4, 6, 7, 8) as its sides, we would be missing the dates (03, 05, 09, 30). This suggests that we must also have a 0 on die 2, so that 0 can occur in combination with all nine other numerals.

In the same vein, although slightly easier to see, we must have both a 1 and a 2 on each of the dice, because 11 and 22 are both valid dates.

 

Second Thoughts:

After having placed 0, 1, and 2 on both dice, we are left with six remaining slots. Our immediate reaction is probably confusion, because we have seven remaining numerals to place. We might be tempted to doubt our earlier conclusions, but our reasoning turns out to be sound, so we are forced to find another way forward.

The pigeonhole principle (see https://en.wikipedia.org/wiki/Pigeonhole_principle) suggests that we may have to fit two of our remaining choices onto one face of a single die. Following this idea through, we see that the most obvious candidates for this are the 6 and the 9, which can be transformed into each other by being flipped.

 

Final Thoughts:

Our task has boiled down to placing the digits 3-8 on the remaining six facets of the two dice. Of these, 3 is the only digit that might give us cause for concern, as it can appear as both the first and second digits of a calendar date (23 and 31, for example). However, since we only need to create the calendar dates, we see that the only two days that begin with 3 are the 30^th and 31^st. Since each of the dice has both a 0 and a 1, however, we see that it is not an issue. We are free to place each of the digits in any of the remaining positions. One possible solution is (012345) and (012678).

 

Going Further:

Since we had some choice in assigning locations to the last 6 digits, one might be tempted to ask how many possible solutions there are to this problem. The calculation is straightforward. Since there are 3 slots on each of the two dice and 6 digits to choose from, a choice of 3 digits to fit on one of the dice will determine the three on the other. Since order does not matter, there are 6 choose 3 = 20 different ways to do this. However, since our situation is symmetrical (each die starts only with faces of 0, 1, and 2), we see that it doesn’t matter which die we put our selection on. It follows that we have to divide our answer by 2, giving a total of 10 possible configurations.