= Solution
A Hermitian two-by-two <matrix> has the form
$$
A=\begin{pmatrix}\alpha&z\\\bar z&\delta\end{pmatrix},\qquad
\alpha,\delta\in\mathbb R,\quad z\in\mathbb C.
$$
Addition and real scalar multiplication preserve this condition, and the two diagonal entries plus the real and imaginary parts of $z$ give four independent real coordinates. Thus \b[$S$ is a real four-dimensional vector space]. The <trace> product is real because $\overline{\operatorname{tr}(AB)}=\operatorname{tr}(BA)=\operatorname{tr}(AB)$; this also proves symmetry of $b$.
Directly, $\operatorname{tr}(A^2)=\alpha^2+\delta^2+2|z|^2$ and $(\operatorname{tr}A)^2=\alpha^2+2\alpha\delta+\delta^2$. Hence
$$
\boxed{b(A,A)=|z|^2-\alpha\delta=-\det A.}
$$
Since $\operatorname{tr}I=2$ and $\operatorname{tr}(AI)=\operatorname{tr}A$,
$$
\boxed{b(A,I)=-\frac12\operatorname{tr}A.}
$$
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