(a) Let the given two-digit number be $\overline{ab}$. The only number that can be obtained by changing the order of digits is $\overline{ba}$. The difference of these numbers is $(10a + b) - (10b + a) = 9(a - b)$. To obtain the largest difference, $a$ must be as large as possible and $b$ as small as possible. Since $b = 0$ is impossible, we must have $a = 9$ and $b = 1$ giving $a - b = 8$. Hence the desired largest difference is $72$.

(b) Let the given three-digit number be $\overline{abc}$. Interchanging the last two digits can change it by less than $100$. Interchanging the first two digits can change the number by at most $720$ by part (a) of the problem. It remains to study cases where changing the order of digits results in $\overline{cab}$, $\overline{bca}$ or $\overline{cba}$.

* The difference of numbers $\overline{abc}$ and $\overline{cab}$ is $(100a + 10b + c) - (100c + 10a + b) = 9(10a + b - 11c)$. To obtain the largest difference, $a$ and $b$ must be as large as possible and $c$ as small as possible. Since $c$ is the first digit of the number, $c = 0$ is impossible, whence the largest difference is obtained if $a = b = 9$ and $c = 1$. This difference is $991 - 199 = 792$.

* The difference of numbers $\overline{abc}$ and $\overline{bca}$ is $(100a + 10b + c) - (100b + 10c + a) = 9(11a - 10b - c)$. To obtain the largest difference, $a$ must be as large as possible and both $b$ and $c$ as small as possible, i.e., $a = 9$, $b = 1$ and $c = 0$. This difference is $910 - 109 = 801$.

* The difference of numbers $\overline{abc}$ and $\overline{cba}$ is $(100a + 10b + c) - (100c + 10b + a) = 99(a - c)$. To obtain the largest difference, $a$ must be as large as possible and $c$ as small as possible. Since $c \neq 0$, the largest difference is obtained if $a = 9$ and $c = 1$. Then the difference of the three-digit numbers is $99 \cdot 8 = 792$.

Consequently, the desired largest difference is $801$.