Solution:

a. Let the digits of the original number be $a_1$, $a_2$, ... and the rearranged digits be $b_1$, $b_2$, ... Suppose that in the addition there is a carry, in other words $a_1 + b_1 > 9$ for some $i$. Take the largest such $i$. Then the resulting digit in that position cannot be a $9$. Contradiction. So there cannot be any carries. Hence each pair $a_1 + b_1 = 9$. Let $n$ be the total number of digits $0, 1, 2, 3,$ and $4$ in the number. Then each of these must be paired with a digit $5, 6, 7, 8$ or $9$. So the total number of digits $5, 6, 7, 8$ and $9$ must also be $n$, and hence the number must have an even number of digits. But we are told that the answer and hence the original number has an odd number of digits.

b. In the addition the carry can never be $2$, because that would require the previous carry to be at least $2$, and the first carry cannot be $2$. So all carries are $0$ or $1$. If a carry is $1$, then all subsequent carries must also be $1$. If the first carry is $0$, then the corresponding digits must be $0$ and hence the original number is divisible by $10$. If it is not, then all carries are $1$ and hence after the first carry all the digit pairs sum to $9$. But arguing as in (a), this means that there must be an even number of digits, excluding the last (where we have a digit sum $10$), and hence an odd number of digits in the original number. But $10^{10}$ has an odd number of digits and hence the original number had an even number of digits. Contradiction.