Since point A $(1,1)$ and point B $(0,-2)$ lie on opposite sides of the line $x+ay+1=0$, the expressions evaluated at these points must have opposite signs.

For point A $(1,1)$, substituting into the line equation gives:
$$(1) + a(1) + 1 > 0$$
$$(a+2) > 0$$

For point B $(0,-2)$, substituting into the line equation gives:
$$(0) + a(-2) + 1 0$$

To find the solution for $a$, we consider the critical points where the expression equals zero, i.e., $a=-2$ and $a=\frac{1}{2}$. We create a sign chart around these critical points:

```
-∞ -2 1/2 ∞
(a+2) - 0 + +
____________
(-2a+1) + - 0 -

(a+2)(-2a+1) - + - +
```

From the chart, we find that the expression $(a+2)(-2a+1)$ is positive when $a \frac{1}{2}$, hence the possible range for $a$ is:
$$a \in (-\infty, -2) \cup (\frac{1}{2}, +\infty)$$

Therefore, the correct choice is:
$\boxed{D (-\infty,}$ -2) ${\cup (\frac{1}{2}, +\infty)}.$