Lexicographic Order

Firstly arrange them in alphabetic order: $\{A,F,H,I,M,T\}$

Now all the permutations when the first letter is A $= 6! = 720$

All the permutations when the first letter is F $= \dfrac{6!}{2!} = 360$ (since A is repeated twice)

All the permutations when the first letter is H $= \dfrac{6!}{2!} = 360$ (since A is repeated twice)

All the permutations when the first letter is I $= \dfrac{6!}{2!} = 360$ (since A is repeated twice)

All the permutations when the first letter is M $= \dfrac{6!}{2!} = 360$ (since A is repeated twice)

All the permutations when the first letter is T $= \dfrac{6!}{2!} = 360$ (since A is repeated twice)

Total words up to H $= 1440$

$1441$ st word = IAAFHMT

We know that the 1443rd word will be (obviously) $IAA****$ . Now we can label $F=1, H=2, M=3, T=4$

And now the question simplifies into arranging 4-digit numbers made by $1,2,3,4$ in ascending order and taking the 4th one.

1: $1234$

2: $1243$

3: $1324$

4: $1342$

Now we can change the numbers back to the corresponding letters, and we will get $IAAFMTH$ . $\ _\square$

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Credit to puffles for the first half of the solution.