2. Simplify the expression $\left(\frac{2^{x}+2^{-x}}{2}\right)^{2}-\left(\frac{2^{x}-2^{-x}}{2}\right)^{2}$.
2. We square the first fraction $\frac{2^{2 x}+2 \cdot 2^{x} \cdot 2^{-x}+2^{-2 x}}{4}$, and similarly the second $\frac{2^{2 x}-2 \cdot 2^{x} \cdot 2^{-x}+2^{-2 x}}{4}$. We subtract the fractions and get $\frac{4 \cdot 2^{x} \cdot 2^{-x}}{4}=1$.

I. Method

In the expression, there is a difference of squares, so we can factor it as

$$
\left(\frac{2^{x}+2^{-x}}{2}+\frac{2^{x}-2^{-x}}{2}\right)\left(\frac{2^{x}+2^{-x}}{2}-\frac{2^{x}-2^{-x}}{2}\right)
$$

In the first parenthesis, we get $\frac{2 \cdot 2^{x}}{2}=2^{x}$, and in the second, $\frac{2 \cdot 2^{-x}}{2}=2^{-x}$. Therefore, the value of the expression is $2^{x} \cdot 2^{-x}=1$.
Squaring the first parenthesis: $\frac{2^{2 x}+2 \cdot 2^{x} \cdot 2^{-x}+2^{-2 x}}{4}$ ..... 1 point
Squaring the second parenthesis: $\frac{2^{2 x}-2 \cdot 2^{x} \cdot 2^{-x}+2^{-2 x}}{4}$ ..... 1 point
Considering the sign before the second fraction ..... 1 point
Subtracting the fractions $4 \cdot 2^{x} \cdot 2^{-x}=4$. ..... $1+1^{*}$ points
Result 1 ..... 1 point
II. Method
Factoring the expressions according to the rule of sum and difference of squares for each factor:
$\frac{2^{x}+2^{-x}}{2}+\frac{2^{x}-2^{-x}}{2}$ ..... 1 point
$\frac{2^{x}+2^{-x}}{2}-\frac{2^{x}-2^{-x}}{2}$ ..... 1 point
Adding in each parenthesis: $\frac{2 \cdot 2^{x}}{2}=2^{x}$ ..... 1 point
$\frac{2 \cdot 2^{-x}}{2}=2^{-x}$ ..... 1 point
Calculating $\frac{2 \cdot 2^{x}}{2} \cdot \frac{2 \cdot 2^{-x}}{2}=2^{x} \cdot 2^{-x}$ ..... 1 point
Result 1 ..... 1 point