Combinatorial Mathematics

well, for your initial stated problem, i would actually just brute-force it on paper- because the time spent doing that is probably distinctly less than the time it takes finding a relevant formula. and it's a pretty specific issue for there to be a tool that would handle it, but that doesn't mean there isn't one somewhere.

that said, http://www.anydice.com is pretty helpful for a lot of similar game-design uses, though not that specific one.

On the question of dice faces adding to 12:
I had a go at this but I don't think the answer is 49 (unless I've misunderstood the question). I spent some time on the problem, and while I haven't arrived at a solution it looks like the number would be in the order of a few hundred combinations. For example, using only 3 faces (while all the other faces stay at zero), I calculated more than 100 combinations.

On the question of a tool to calculate these things:
Any spreadsheet application.

On the question of the roads on tiles:
The first phrasing of the question would lead to an enormous number of potential tiles (not to mention that many of them would be impractical to draw on a normal-sized hex tile). Without doing the maths, I think the solution would number into many thousands of unique tiles.
The second phrasing of the question is simpler, and also more practical to draw. The number of combinations would be near to but less than 3^6. The reason it would be less is that a hex can be rotated without becoming unique. (For example, the 6 combinations of "1 dead end" are actually the same tile.)

As an aside:
I originally misinterpreted your description of the tiles and calculated a solution for something you didn't ask. But I'll describe my assumptions anyway and then the solution. (I've seen a game that used hex tiles almost exactly as per your second solution, and while the mechanics worked fine I didn't like the visual appeal because it created jagged corners rather than curves.)
- A tile can have at most one dead-end. (I pictured this as terminating in the middle of the tile, so if there were two "dead-ends" they would meet in the middle and thus become a road.)
- A side can have at most one road. (I pictured this as curving between the two sides, without passing through the middle.)
- The roads cannot cross.
Under these assumptions, there are only 25 possible tiles.
- 1x one dead-end
- 3x one road
- 8x two roads
- 2x three roads
- 8x one dead-end and one road
- 3x one dead-end and two roads
I may have missed a couple, but you can see the final number is still quite small.
If multiple roads can connect to the same side, the combinations increase substantially.

Regards,
kos

I stand corrected. I was trying to come up with a generalised formula to solve the problem for any size dice with any target number, but obviously there was an error in my calculations somewhere.

Note that maximum for "option 2" would be 3^6 (729) not 6^3 (216).
So the solution would be somewhere between 121 and 729 depending on how many of the tiles are rotations of each other. If an approximation is sufficient for your needs, you could just take the mid-point of 425.

*But* ... I just realised this number doesn't take into account multiple roads connecting to each side. With multiple roads, this number would go up again. Actually calculating the number would get very difficult and further complicated by rotations.

For example, a tile with 3 roads terminating on sides 1, 3, and 5 would have 2 unique combinations (all 3 sides connected; any 2 sides connected (the others are rotations of this)). However, a tile with 3 roads terminating on sides 1, 2, and 4 would have 4 unique combinations (all 3 sides connected; 1-2 and 1-4; 1-2 and 2-4; 1-4 and 2-4).

At a rough guess I'd put the total number between 1000 and 2000, after applying fudge factors for the rotations and the multiple roads connecting to the same side.

Regards,
kos