Suppose $a$ and $b$ are real numbers such that \[\lim_{x\to 0}\frac{\sin^2 x}{e^{ax}-bx-1}=\frac{1}{2}.\] Determine all possible ordered pairs $(a, b)$.
1. Given the limit:
   \[
   \lim_{x\to 0}\frac{\sin^2 x}{e^{ax}-bx-1}=\frac{1}{2}
   \]
   We recognize that this is an indeterminate form of type \(\frac{0}{0}\) as \(x \to 0\).

2. Apply L'Hôpital's Rule:
   \[
   \lim_{x\to 0}\frac{\sin^2 x}{e^{ax}-bx-1} = \lim_{x\to 0}\frac{2\sin x \cos x}{ae^{ax} - b}
   \]
   Simplify the numerator using the double-angle identity \(\sin(2x) = 2\sin x \cos x\):
   \[
   \lim_{x\to 0}\frac{\sin(2x)}{ae^{ax} - b}
   \]

3. Evaluate the limit as \(x \to 0\):
   - The numerator \(\sin(2x) \to 0\).
   - The denominator \(ae^{ax} - b \to a - b\).

   If \(a \neq b\), the limit would be:
   \[
   \lim_{x\to 0}\frac{0}{a - b} = 0
   \]
   This contradicts the given limit of \(\frac{1}{2}\). Therefore, we must have \(a = b\).

4. Substitute \(a = b\) and apply L'Hôpital's Rule again:
   \[
   \lim_{x\to 0}\frac{\sin^2 x}{e^{ax} - ax - 1} = \lim_{x\to 0}\frac{2\sin x \cos x}{ae^{ax} - a}
   \]
   Simplify the numerator using the double-angle identity:
   \[
   \lim_{x\to 0}\frac{\sin(2x)}{ae^{ax} - a}
   \]

5. Apply L'Hôpital's Rule again:
   \[
   \lim_{x\to 0}\frac{2\cos(2x)}{a^2e^{ax}}
   \]

6. Evaluate the limit as \(x \to 0\):
   - The numerator \(2\cos(2x) \to 2\).
   - The denominator \(a^2e^{ax} \to a^2\).

   Therefore:
   \[
   \lim_{x\to 0}\frac{2}{a^2} = \frac{1}{2}
   \]

7. Solve for \(a\):
   \[
   \frac{2}{a^2} = \frac{1}{2} \implies a^2 = 4 \implies a = \pm 2
   \]

8. Since \(a = b\), the possible ordered pairs \((a, b)\) are:
   \[
   (2, 2) \quad \text{and} \quad (-2, -2)
   \]

The final answer is \(\boxed{(2, 2), (-2, -2)}\)