= Solution
Fix $T>0$. The given <normal distribution> and part (d) imply
$$
\mathbb E e^{-\lambda\langle X\rangle_T}=e^{-\lambda T}\qquad(\lambda\ge0).
$$
One can deduce determinism without any moment assumption on the bracket. Put $Z=e^{-\langle X\rangle_T}$. Taking $\lambda=1,2$ gives $\mathbb EZ=e^{-T}$ and $\mathbb EZ^2=e^{-2T}$, so $\operatorname{Var}(Z)=0$. Therefore $\langle X\rangle_T=T$ almost surely. Applying this at every rational time and using continuity of <quadratic variation> gives $\langle X\rangle_t=t$ simultaneously for all $t\ge0$ outside a single null set.
The <Lévy characterization of Brownian motion> states that a continuous local martingale starting at zero with this bracket is <Brownian motion> in its filtration. To see the independent-increment conclusion directly, the <Itô formula> shows that $e^{i\theta X_t+\theta^2t/2}$ is a martingale on any fixed bounded time interval: it is a local martingale with a deterministic bound on its modulus. Thus
$$
\mathbb E[e^{i\theta(X_t-X_s)}\mid\mathcal F_s]=e^{-\theta^2(t-s)/2}.
$$
The deterministic <conditional characteristic function> identifies an $N(0,t-s)$ increment independent of $\mathcal F_s$. Together with the given path continuity and $X_0=0$, this proves \b[$X$ is Brownian motion].
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