Brazilian Math Olympiad

a. The largest sum is obtained when the largest digits are assigned to the leftmost positions, so it is equal to $(8+7) \cdot 1000 + (6+5) \cdot 100 + (4+3) \cdot 10 + 2 + 1 = 16373$.

b. The smallest sum is obtained when the smallest digits are assigned to the leftmost positions. So $1$ and $2$ are in the thousands, $3$ and $4$ are in the hundreds, $5$ and $6$ are in the tens and $7$ and $8$ are in the units. The biggest of the two numbers is the one beginning with $2$. The minimum of its difference is obtained by making the digits in the smallest number bigger and the digits in the biggest number smaller, so it is $2357 - 1468 = 889$.