= Solution
Putting the <unit> vector along a coordinate axis gives $\sigma_i^2=I$. Putting it along $(e_i+e_j)/\sqrt2$ gives $\sigma_i\sigma_j+\sigma_j\sigma_i=0$ for $i\ne j$. Combining the resulting anticommutators with the given <commutators> yields $\sigma_i\sigma_j=\delta_{ij}I+i\epsilon_{ijk}\sigma_k$, and hence
$$
\boxed{(\mathbf a\cdot\boldsymbol\sigma)(\mathbf b\cdot\boldsymbol\sigma)
=(\mathbf a\cdot\mathbf b)I+i(\mathbf a\times\mathbf b)\cdot\boldsymbol\sigma.}
$$
The operators $S_i=(\hbar/2)\sigma_i$ obey angular-momentum <commutators> and $\mathbf S^2=3\hbar^2I/4$, the spin-one-half value. In a spin-half irreducible sector they are the intrinsic <spin> operators; the algebra also permits direct sums of such sectors.
Since $Q=\mathbf n\cdot\boldsymbol\sigma$ is Hermitian with $Q^2=I$, its operators $P_\pm=(I\pm Q)/2$ are Hermitian and satisfy $P_\pm^2=P_\pm$, $P_+P_-=0$ and $P_++P_-=I$. Moreover $QP_\pm=\pm P_\pm$. Thus $|\chi\rangle=P_+|\chi\rangle+P_-|\chi\rangle$ is its <orthogonal> <eigenstate> decomposition, omitting any zero component. The Born probabilities for a normalized state are
$$
\boxed{\|P_\pm|\chi\rangle\|^2=\langle\chi|P_\pm|\chi\rangle=\tfrac12(1\pm\langle\chi|Q|\chi\rangle).}
$$
For <spin> up along $\mathbf m$, the anticommutator $\{\mathbf m\cdot\boldsymbol\sigma,\sigma_j\}=2m_jI$ gives $\langle\sigma_j\rangle=m_j$. Therefore \b[the probability of <spin> up along $\mathbf n$ is $(1+\mathbf n\cdot\mathbf m)/2$].
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