If `lim_(xto0)(f(x))/(sin^(2)x)=8,lim_(xto0)(g(x))/(2cosx-xe^(x)+x^(3)+x-2)=lamda" and "`
`lim_(x to 0)(1+2f(x))^((1)/(g(x)))=(1)/(e)," then"`
`lim_(x to 0)(1+f(x))^((1)/(2g(x)))` is equal to

Explanation

To solve the given problem step by step, we will analyze the limits provided and derive the necessary expressions.

### Step 1: Analyze the first limit
We are given:
\[ \lim_{x \to 0} \frac{f(x)}{\sin^2 x} = 8 \]
As \( x \to 0 \), \(\sin^2 x \to 0\). For the limit to be finite, \(f(x)\) must also approach 0. Thus, we can conclude that:
\[ \lim_{x \to 0} f(x) = 0 \]
Now, we can rewrite the limit:
\[ \lim_{x \to 0} \frac{f(x)}{x^2} \cdot \frac{x^2}{\sin^2 x} = 8 \]
Using the fact that \(\lim_{x \to 0} \frac{\sin x}{x} = 1\), we have:
\[ \lim_{x \to 0} \frac{x^2}{\sin^2 x} = 1 \]
Thus:
\[ \lim_{x \to 0} \frac{f(x)}{x^2} = 8 \]

### Step 2: Analyze the second limit
Next, we consider:
\[ \lim_{x \to 0} \frac{g(x)}{2 \cos x - x e^x + x^3 + x - 2} = \lambda \]
As \(x \to 0\), we can simplify the denominator:
\[ 2 \cos x - 2 = 2(\cos x - 1) \approx -2 \cdot \frac{x^2}{2} = -x^2 \quad \text{(using Taylor expansion)} \]
For the term \(x e^x\), we have:
\[ x e^x \approx x(1 + x) = x + x^2 \]
Thus:
\[ 2 \cos x - x e^x + x^3 + x - 2 \approx -x^2 - (x + x^2) + x + x^3 \approx -2x^2 + x^3 \]
So the limit becomes:
\[ \lim_{x \to 0} \frac{g(x)}{-2x^2} = \lambda \]
This implies:
\[ \lim_{x \to 0} g(x) = -2\lambda \]

### Step 3: Analyze the third limit
We have:
\[ \lim_{x \to 0} (1 + 2f(x))^{\frac{1}{g(x)}} = \frac{1}{e} \]
This is a \(1^\infty\) form, so we can rewrite it using the exponential limit:
\[ \lim_{x \to 0} \frac{2f(x)}{g(x)} = -1 \]
Substituting the limits we found:
\[ \frac{2 \cdot 8}{-2\lambda} = -1 \implies \frac{16}{-2\lambda} = -1 \implies \lambda = 8 \]

### Step 4: Find the final limit
Now we need to find:
\[ \lim_{x \to 0} (1 + f(x))^{\frac{1}{2g(x)}} \]
This is again a \(1^\infty\) form, so we can rewrite it:
\[ \lim_{x \to 0} (1 + f(x))^{\frac{1}{2g(x)}} = e^{\lim_{x \to 0} \frac{f(x)}{2g(x)}} \]
Substituting the limits:
\[ \lim_{x \to 0} \frac{f(x)}{2g(x)} = \frac{8}{2 \cdot (-16)} = \frac{8}{-32} = -\frac{1}{4} \]
Thus:
\[ \lim_{x \to 0} (1 + f(x))^{\frac{1}{2g(x)}} = e^{-\frac{1}{4}} = \frac{1}{e^{1/4}} \]

### Final Answer
\[ \lim_{x \to 0} (1 + f(x))^{\frac{1}{2g(x)}} = e^{-\frac{1}{4}} \]