We are solving the problem:

$\lim_{x \to 0} (\cos x)^{\frac{1}{\sin^2 x}}.$

Step-by-Step Solution:

1. Rewrite the Expression Using Logarithms
   Let $y = (\cos x)^{\frac{1}{\sin^2 x}}$.
   Taking the natural logarithm on both sides: $\ln y = \frac{1}{\sin^2 x} \ln(\cos x).$

2. Simplify the Exponent
   The goal is to evaluate $\ln y$ as $x \to 0$. Since $\ln y = \frac{\ln(\cos x)}{\sin^2 x}$, we focus on the limit: $\lim_{x \to 0} \frac{\ln(\cos x)}{\sin^2 x}.$

3. Expand $\ln(\cos x)$ for Small $x$
   Using the approximation $\cos x \approx 1 - \frac{x^2}{2}$ for small $x$, we get: $\ln(\cos x) \approx \ln\left(1 - \frac{x^2}{2}\right).$ For small $x$, $\ln(1 - u) \approx -u$, so: $\ln(\cos x) \approx -\frac{x^2}{2}.$

4. Substitute Back into the Limit
   Substitute $\ln(\cos x) \approx -\frac{x^2}{2}$ into the limit: $\lim_{x \to 0} \frac{\ln(\cos x)}{\sin^2 x} \approx \lim_{x \to 0} \frac{-\frac{x^2}{2}}{\sin^2 x}.$ Using the approximation $\sin x \approx x$ for small $x$, we have $\sin^2 x \approx x^2$. Thus: $\lim_{x \to 0} \frac{-\frac{x^2}{2}}{\sin^2 x} \approx \lim_{x \to 0} \frac{-\frac{x^2}{2}}{x^2} = -\frac{1}{2}.$

5. Interpret the Result
   Since $\ln y \to -\frac{1}{2}$, we have: $y = e^{\ln y} \to e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}}.$

Final Answer:

$\boxed{\frac{1}{\sqrt{e}} \text{ (Option B)}}$

Would you like further details or have questions? Here are five related questions to expand on this concept:

1. What are the key approximations used for small values of $x$ in trigonometric and logarithmic functions?
2. How does the substitution $\ln(\cos x) \approx -\frac{x^2}{2}$ simplify the problem?
3. Can you explain why $\sin x \approx x$ is valid for small $x$?
4. How would the solution differ if the power were $1/\sin x$ instead of $1/\sin^2 x$?
5. How can L’Hôpital’s Rule be used to solve limits like this one?

Tip: For problems involving small $x$, always expand functions into their Taylor series to simplify the analysis.