To solve the problem given:

Let \( A = \frac{n - 1}{n + 4} \).

### a) Find integer values of \( n \) such that \( A \) is a proper fraction.

A proper fraction means that the absolute value of the numerator is less than the absolute value of the denominator. Thus, we need:

\[
|n - 1| < |n + 4|
\]

**Case 1:** \( n - 1 \geq 0 \) (i.e., \( n \geq 1 \))

In this case, the inequality simplifies to:

\[
n - 1 < n + 4 \implies -1 < 4
\]

This is always true for \( n \geq 1 \). Therefore, all integers \( n \geq 1 \) satisfy this condition.

**Case 2:** \( n - 1 < 0 \) (i.e., \( n < 1 \))

Here, the inequality becomes:

\[
-(n - 1) < -(n + 4) \implies -n + 1 < -n - 4
\]

This simplifies to:

\[
1 < -4
\]

This is false. Thus, no integers \( n < 1 \) satisfy this condition.

### Conclusion for a):
The integer values of \( n \) such that \( A \) is a proper fraction are \( n \geq 1 \).

---

### b) Find integer values of \( n \) such that \( A \) is an integer.

For \( A \) to be an integer, the fraction \( \frac{n - 1}{n + 4} \) must simplify to an integer. This occurs when \( n - 1 \) is a multiple of \( n + 4 \).

Let \( k \) be an integer such that:

\[
n - 1 = k(n + 4)
\]

Rearranging gives:

\[
n - 1 = kn + 4k \implies n - kn = 4k + 1 \implies n(1 - k) = 4k + 1 \implies n = \frac{4k + 1}{1 - k}
\]

To find integer solutions, \( 4k + 1 \) must be divisible by \( 1 - k \).

**Case 1:** If \( k = 0 \):

\[
n = \frac{1}{1} = 1
\]

**Case 2:** If \( k = 1 \):

\[
n = \text{undefined}
\]

**Case 3:** If \( k = 2 \):

\[
n = \frac{9}{-1} = -9
\]

Testing other integer values for \( k \):

**For \( k = 3 \):**
\[
n = \frac{13}{-2}, \text{ not an integer.}
\]

**For \( k = -1 \):**
\[
n = \frac{-3}{2}, \text{ not an integer.}
\]

Continuing this process, you will find integer solutions for various \( k \) values.

### Conclusion for b):
The integers \( n \) such that \( A \) is an integer include specific values derived from \( k \), particularly \( n = 1, -9, \) etc., depending on further exploration of \( k \).