Since this was solved for non-distinct digits, I'm trying to get the maximum out of "MATH" with distinct digits.
So the first try is with

M = 9. We imediately see that S needs to be 8. We get

  AWE +
 8O9E =
 9ATH
 

Since the hundreds of "AWE" and "MATH" need to be the same ("A") ?this means that "O" needs to be 0 and 9+W must not go over 10 so we don't get a 1 to carry. this means that W is also 0. So this does not work.

So my seconds try is with

M = 8 so S = 7

  AWE +
 7O8E =
 8ATH
 

Now we try to maximize "A". Let's go with the obvious 9. This means again O = 0 and we get.

  9WE +
 708E =
 89TH
 

now 8+w must not go over 10 so w = 1 ('cuz 0 is taken). But this leads to T being either 9 or 0 (if E+E go over 10) but both 9 and 0 are taken.
continuing to maximize A.... so 9 does not work, 8 and 7 are taken by M and S....so 6

  6WE +
 7O8E =
 86TH
 

if O = 0 we end up with a similar contradiction as above, so lets try with o = 9 and w+8+(a possible carry from E+E) go over 10.

  6WE +
 798E =
 86TH
 

Now lets maximize T in math. 9, 8, 7, 6 are taken so lets go with 5. This means W+8+(possible carry from E+E) = 15. This leads to W being either 6 or 7 but they are already taken by A and S.
so lets go with T = 4. This means W+8+(possible carry from E+E) = 14. So W is 5 or 6. 6 is taken by A so W = 5.

  65E +
 798E =
 864H
 

NOw E+ E needs to go over 10. but all of 5, 6, 7, 8, 9 are taken. So this is no good.
so let's try with T = 3. This means W+8+(possible carry from E+E) = 13. So W is 4 or 5. If w = 5

  65E +
 798E =
 863H
 

E+E should be below 10. E cannot be 0 because H would be 0. SO it can be 1, 2, 4 (3 is taken by T). If it's 4 then H would be 8 which is taken by M. It works with both 1 and 2 but to maximize "math" we go with 2.

So the final result:

  652 +
 7982 =
 8634
 

Since this was solved for non-distinct digits, I'm trying to get the maximum out of "MATH" with distinct digits.
So the first try is with

M = 9. We imediately see that S needs to be 8. We get

  AWE +
 8O9E =
 9ATH
 

Since the hundreds of "AWE" and "MATH" need to be the same ("A") ?this means that "O" needs to be 0 and 9+W must not go over 10 so we don't get a 1 to carry. this means that W is also 0. So this does not work.

So my seconds try is with

M = 8 so S = 7

  AWE +
 7O8E =
 8ATH
 

Now we try to maximize "A". Let's go with the obvious 9. This means again O = 0 or 1 and we get.

  9WE +
 708E =
 89TH
 

now 8+w must not go over 10 so w = 1 ('cuz 0 is taken). But this leads to T being either 9 or 0 (if E+E go over 10) but both 9 and 0 are taken.
continuing to maximize A.... so 9 does not work, 8 and 7 are taken by M and S....so 6

  6WE +
 7O8E =
 86TH
 

if O = 0 we end up with a similar contradiction as above, so lets try with o = 9 and w+8+(a possible carry from E+E) go over 10.

  6WE +
 798E =
 86TH
 

Now lets maximize T in math. 9, 8, 7, 6 are taken so lets go with 5. This means W+8+(possible carry from E+E) = 15. This leads to W being either 6 or 7 but they are already taken by A and S.
so lets go with T = 4. This means W+8+(possible carry from E+E) = 14. So W is 5 or 6. 6 is taken by A so W = 5.

  65E +
 798E =
 864H
 

NOw E+ E needs to go over 10. but all of 5, 6, 7, 8, 9 are taken. So this is no good.
so let's try with T = 3. This means W+8+(possible carry from E+E) = 13. So W is 4 or 5. If w = 5

  65E +
 798E =
 863H
 

E+E should be below 10. E cannot be 0 because H would be 0. SO it can be 1, 2, 4 (3 is taken by T). If it's 4 then H would be 8 which is taken by M. It works with both 1 and 2 but to maximize "math" we go with 2.

So the final result:

  652 +
 7982 =
 8634