Using four different digits, form the subtraction 5A - 4B to make the answer as small as possible and as large as possible.

To make the subtraction $(50+A) - (40+B) = 10 + A - B$ as small as possible, we need to choose the smallest possible digit for $A$ and the largest possible digit for $B$. The digits 5 and 4 are already used, so we cannot use them for $A$ or $B$. The available digits from 1 to 9 (excluding 4 and 5) are $\{1, 2, 3, 6, 7, 8, 9\}$.

• Smallest $A$: Choose $A=1$.
• Largest $B$: Choose $B=9$. The four different digits used are $5, 1, 4, 9$.

Substitute $A=1$ and $B=9$ into the subtraction:

$$\begin{array}{c} 5 \quad 1 \\ - 4 \quad 9 \\ \hline \end{array}$$ $$51 - 49 = 2$$

The completed subtraction for the smallest answer is:

$$\boxed{ \begin{array}{c} 5 \quad 1 \\ - 4 \quad 9 \\ \hline 0 \quad 2 \end{array} }$$

To make the subtraction $(50+A) - (40+B) = 10 + A - B$ as large as possible, we need to choose the largest possible digit for $A$ and the smallest possible digit for $B$. Again, the digits 5 and 4 are already used. The available digits from 1 to 9 (excluding 4 and 5) are $\{1, 2, 3, 6, 7, 8, 9\}$.

• Largest $A$: Choose $A=9$.
• Smallest $B$: Choose $B=1$. The four different digits used are $5, 9, 4, 1$.

Substitute $A=9$ and $B=1$ into the subtraction:

$$\begin{array}{c} 5 \quad 9 \\ - 4 \quad 1 \\ \hline \end{array}$$ $$59 - 41 = 18$$

The completed subtraction for the largest answer is:

$$\boxed{ \begin{array}{c} 5 \quad 9 \\ - 4 \quad 1 \\ \hline 1 \quad 8 \end{array} }$$