= Solution
Taking $n=e_i$ gives $\sigma_i^2=I$. Taking $n=(e_i+e_j)/\sqrt2$ for $i\ne j$ and expanding the square gives $\sigma_i\sigma_j+\sigma_j\sigma_i=0$. Thus the <anticommutator> and <commutator> together imply
$$
\sigma_i\sigma_j=\delta_{ij}I+i\epsilon_{ijk}\sigma_k,
\qquad\boxed{(a\cdot\sigma)(b\cdot\sigma)=(a\cdot b)I+i(a\times b)\cdot\sigma.}
$$
Since $n\cdot\sigma$ is <Hermitian> and $\theta$ is real, the exponent is <anti-Hermitian>. Its exponential has inverse equal to its adjoint and is therefore <unitary>. Splitting the <exponential series> into even and odd powers, using $(n\cdot\sigma)^2=I$, gives
$$
\boxed{U=I\cos(\theta/2)-i(n\cdot\sigma)\sin(\theta/2).}
$$
Put $N=n\cdot\sigma,M=m\cdot\sigma$ with $m\perp n$. Then $NM=i(n\times m)\cdot\sigma$, $MN=-NM$, and $NMN=-M$. Expanding $(c-iNs)M(c+iNs)$ gives $(c^2-s^2)M-2icsNM$. Consequently
$$
\boxed{UMU^{-1}=(m\cdot\sigma)\cos\theta+((n\times m)\cdot\sigma)\sin\theta.}
$$
The spin-half <angular momentum> is $S_i=\hbar\sigma_i/2$. The commutators then become $[S_i,S_j]=i\hbar\epsilon_{ijk}S_k$, and $(n\cdot S)^2=\hbar^2I/4$ expresses the two possible directional <spin> values $\pm\hbar/2$. Also $S^2=3\hbar^2I/4$, appropriate to <spin> one half. The exponential represents a spatial rotation on <spin> states.
Choose $n=(0,1,0)$ and initially $\sigma_3|\chi\rangle=|\chi\rangle$. The conjugation formula sends $\sigma_3$ to $\sigma_3\cos\theta+\sigma_1\sin\theta$. Therefore
$$
(\sigma_1\sin\theta+\sigma_3\cos\theta)U|\chi\rangle
=U\sigma_3|\chi\rangle=U|\chi\rangle.
$$
Thus \b[$U|\chi\rangle$ is <spin> up along $(\sin\theta,0,\cos\theta)$], as required.
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