For example, if I am rolling two dice, one with 8 faces and the other with 10. How do I calculate...

...the odds of rolling a certain combo, say, a 1 and a 10,
...the odds of rolling a certain digit on either die, say a 4
...the odds of the total value, like "the odds of a total of 12", and finally,
...the average value of any given roll?

I'm designing a board game, and so I'm trying to figure some of these combos out.

I can figure the odds of each die, for an 8 and 10 sided, each face is 1/8. But I don't know how to combine odds... like the odds of rolling a 9 could be a 1 and an 8, 2 and a 7, etc. But I get confused with the odds of rolling a 1 on *either* die - they're different, b/c of the different # of faces.

Also average value, I have no clue where to begin.

Thank you!
Probability of getting a 1 either die=

1-probability of getting a 1 on neither die.
Rolling a combo: it compounds multiplicatively. If you want to succeed an exact roll on a d8, it's 1/8. On a d10, it's 1/10. On both a d8 and d10, it's a 1/8 \* 1/10 = 1/80, or 1.25%

Rolling a digit on either: obviously this depends on if the digit is on both dice, but since it's trivial otherwise I'll assume it is. As the other comment said it's the inverse of getting the wrong digit on both dice. In the d8 / d10 case, it would be:

1 - (7/8)(9/10) = 1 - 63/80 = 17/80. or 21.25%

Rolling a given total: you need to count all the possible ways this can happen, then divide by 80 since that's the total number of rolls. (In the general case, divide by the product of the number of faces.) So for a 12, with d8 first and d10 second, you could roll: 2 + 10; 3 + 7; 4 + 8; 5 + 7; 6 + 6; 7 + 5; or 8 + 4. That's 7 possible ways out of 80, so it's 7/80 or 7.85%.

Finally, expectancy can be calculated by simple addition. The average roll of an n-sided dice is (n + 1)/2: for a d8 it's 4.5, for a d10 it's 5.5, so your added roll has an average of 10.