The number is $80$. Charlie was told the sum of digits is $8$. Hanna was told there are $10$ divisors: the possible numbers with $10$ divisors are $48$ and $80$, and both are even, hence Hanna's first statement.
Charlie's candidates for the number are $8, 17, 26, 35, 44, 53, 62, 71, 80$. $8$ has $4$ divisors, but so does $15$; $17$ has $2$ divisors, but so does $2$; and so on for all the other candidates except $80$, which has $10$ divisors, and the only other integer in $\{1,\ldots,99\}$ with $10$ divisors is $48$ which is also even. Hence Charlie's second statement.
Hanna now knows the number can't be $48$, because if Charlie had been told the sum of digits was $12$ he couldn't decide whether the number was $48$ or $66$: $66$ has $8$ divisors, and all possible numbers with $8$ divisors ($24, 30, 40, 42, 54, 56, 66, 70, 78, 88$) are even. Hence Hanna's second statement.
EDIT: This was for a version of the problem where the possible numbers were $1$ to $99$. It is not correct for the version where the possible numbers are $10$ to $99$. I'll post a solution for that version if time permits.
EDIT: Where the possible numbers are $10$ to $99$, the number is not $80$, because after Hanna's first statement Charlie would not know that the number is not $17$ (all the possible numbers with $2$ divisors are odd).
Instead, the number in this case is $30$. Charlie was told the sum of digits is $3$. Hanna was told there are $8$ divisors; all possible numbers with $8$ divisors are even, hence Hanna's first statement.
Now Charlie's candidates for the number are $12, 21, 30$. $12$ has $6$ divisors, as does $45$; $21$ has $4$ divisors, as does $15$; so Charlie knows it must be $30$. Hence Charlie's second statement. The only sums of digits that would have allowed Charlie to make this second statement (without knowing the answer at the beginning statement) are $2$ (with answer $11$), $3$, $14$ (with answer $59$), and $17$ (with answer $89$).
Now Hanna knows the answer is $30$, which is the only number in the intersection of $\{24, 30, 40, 42, 54, 56, 66, 70, 78, 88\}$ and $\{11,30,59,89\}$. Hence Hanna's second statement.
How did I come up with the answer? A process of elimination. After Charlie's first statement the possible numbers are $11,12,\ldots,98$. After Hanna's first the possibilities are restricted to those with $2$, $3$, $8$, $10$, or $12$ divisors ($40$ possibilities). After Charlie's second statement the possibilities are restricted to $11, 30, 59, 89$. And of those all but $30$ have $2$ divisors.