How many ordered pairs of real numbers $ (a, b) $ satisfy equality $\lim_{x \to 0} \frac{\sin^2x}{e^{ax}-2bx-1}= \frac{1}{2}$?
To solve the problem, we need to find the ordered pairs \((a, b)\) such that the limit 

\[
\lim_{x \to 0} \frac{\sin^2 x}{e^{ax} - 2bx - 1} = \frac{1}{2}
\]

is satisfied. 

1. **Initial Setup and Indeterminate Form:**
   We start by noting that as \( x \to 0 \), both the numerator and the denominator approach 0, leading to an indeterminate form \(\frac{0}{0}\). Therefore, we can apply L'Hôpital's Rule.

2. **First Application of L'Hôpital's Rule:**
   Applying L'Hôpital's Rule, we differentiate the numerator and the denominator with respect to \( x \):

   \[
   \lim_{x \to 0} \frac{\sin^2 x}{e^{ax} - 2bx - 1} = \lim_{x \to 0} \frac{2\sin x \cos x}{ae^{ax} - 2b}
   \]

3. **Simplifying the Expression:**
   As \( x \to 0 \), \(\sin x \approx x\) and \(\cos x \approx 1\). Therefore, the limit becomes:

   \[
   \lim_{x \to 0} \frac{2x \cdot 1}{ae^{ax} - 2b} = \lim_{x \to 0} \frac{2x}{ae^{ax} - 2b}
   \]

4. **Second Application of L'Hôpital's Rule:**
   We still have an indeterminate form \(\frac{0}{0}\), so we apply L'Hôpital's Rule again:

   \[
   \lim_{x \to 0} \frac{2x}{ae^{ax} - 2b} = \lim_{x \to 0} \frac{2}{a^2 e^{ax}}
   \]

5. **Evaluating the Limit:**
   As \( x \to 0 \), \( e^{ax} \to 1 \). Therefore, the limit simplifies to:

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

6. **Setting the Limit Equal to \(\frac{1}{2}\):**
   We are given that this limit equals \(\frac{1}{2}\):

   \[
   \frac{2}{a^2} = \frac{1}{2}
   \]

7. **Solving for \(a\):**
   Solving the equation for \(a\):

   \[
   2 \cdot 2 = a^2 \implies a^2 = 4 \implies a = \pm 2
   \]

8. **Conclusion for \(b\):**
   The value of \(b\) does not affect the limit, so \(b\) can be any real number.

Therefore, the ordered pairs \((a, b)\) that satisfy the given limit are \((2, b)\) and \((-2, b)\) for any real number \(b\).

The final answer is \( \boxed{ (2, b) } \) and \((-2, b)\) for any \(b \in \mathbb{R}\).