= Solution
Put $U=M-\tfrac12[M]$. Since the bracket is continuous <finite variation>, $[U]=[M]$. The <Itô formula> applied to $Z=e^U$ gives
$$
dZ_t=Z_t\,dU_t+\frac12Z_t\,d[U]_t=Z_t\,dM_t-\frac12Z_t\,d[M]_t+\frac12Z_t\,d[M]_t.
$$
Therefore
$$
\boxed{dZ_t=Z_t\,dM_t.}
$$
With the printed exponential formula, $Z_0=e^{M_0}$. The usual unit-initial-value <stochastic exponential> is $\exp(M_t-M_0-\tfrac12[M]_t)$; the two coincide when $M_0=0$, as in part (c).
For any strictly positive solution of this stochastic equation, its bracket is $d[Z]_t=Z_t^2d[M]_t$. Applying Itô to the logarithm, after localizing away from zero if necessary, gives
$$
d\log Z_t=\frac{dZ_t}{Z_t}-\frac{d[Z]_t}{2Z_t^2}=dM_t-\frac12d[M]_t.
$$
The same identity holds for $Z'$, so $\log Z_t'-\log Z_t$ is constant in time and is zero initially. Their continuous versions agree at all times on one full-probability event. Thus <pathwise uniqueness for a multiplicative martingale equation> gives
$$
\boxed{\mathbb P(Z_t'=Z_t\text{ for every }t\geq0)=1.}
$$
This is <indistinguishability of stochastic processes>, not just equality almost surely at each separately chosen time.
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