(2) $\left(\frac{1-\mathrm{i}^{\wedge}}{1+\mathrm{i}^{1}}\right)^{\circ}$.

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(2) $\left(\frac{1-\mathrm{i}^{\wedge}}{1+\mathrm{i}^{1}}\right)^{\circ}$.
Correct solution: $\left(\frac{1-i}{1+i}\right)^{5}$
$$
\begin{array}{l}
=\left[\frac{(1-i)^{2}}{(1+i)(1-i)}\right]^{5}=\left(\frac{-2 i}{2}\right)^{5} \\
=(-i)^{5}=-i .
\end{array}
$$