= Solution
Use just the parameters $1$ and $-1$. Positivity and adaptation of their local <martingales> give
$$
X_t=\tfrac12(\log Z_t^{(1)}-\log Z_t^{(-1)}),\qquad A_t=-\log Z_t^{(1)}-\log Z_t^{(-1)}.
$$
Thus both processes are adapted. A positive continuous local <martingale> has a <semimartingale> logarithm, by <Itô formula> after localization away from zero. Since $X=\log Z^{(1)}+A/2$ and $A$ has <finite variation>, $X$ is a <continuous semimartingale>. Write $X=N+V$, where $N$ is a zero-starting <continuous local martingale> and $V$ is continuous <finite variation>, also starting from zero.
Apply <Itô formula> to each exponential:
$$
dZ^{(\theta)}=\theta Z^{(\theta)}\,dN+Z^{(\theta)}\left(\theta\,dV+\tfrac12\theta^2(d[N]-dA)\right).
$$
Because $Z^{(\theta)}$ is itself a local <martingale>, the finite-variation term is zero by part (a). Divide its finite signed measure by the strictly positive $Z^{(\theta)}$. For $\theta=1,-1$, respectively,
$$
dV+\tfrac12(d[N]-dA)=0,\qquad -dV+\tfrac12(d[N]-dA)=0.
$$
Adding and subtracting show $V=0$ and $[N]=A$. Therefore the <exponential criterion for a continuous local martingale and its bracket> gives
$$
\boxed{X\text{ is a continuous local martingale},\qquad [X]=A\text{ indistinguishably}.}
$$
The <semimartingale> property was established before applying Itô's formula to $X$; it was not assumed from the outset.
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