If $(B_t)$ is a Brownian motion and adapted to the filtration $(\mathcal F_t)$ does $B_t-B_s$ is independent of $\mathcal F_s$? Let $(B_t)$ a Brownian motion adapted to the filtration $(\mathcal F_t)$.
Q1) If $t> s$, does $B_t-B_s$ is necessarily independent of $\mathcal F_s$ or $(\mathcal F_t)$ must be the natural filtration of $(B_t)$ to have $B_t-B_s$ independent of $\mathcal F_s$ ?
Q2) Does $(B_t)$ is necessarily a martingale w.r.t. $(\mathcal F_t)$ or $(\mathcal F_t)$ must be the natural filtration for $(B_t)$ being a martingale w.r.t. to $(\mathcal F_t)$ ?

The filtration $(\mathcal{F}_t)_{t \geq 0}$ does not need to be necessarily the natural filtration, but you can't pick just any filtration.
For instance, consider the filtration $$\mathcal{F}_t := \sigma(B_r; r \geq 0).$$ Note that $\mathcal{F}_t$ actually does not depend on $t$. It is trivial that $(B_t)_{t \geq 0}$ is adapted to this filtration. Moreover, you can easily show that $B_t-B_s$ is not independent from $\mathcal{F}_s$ and that $(B_t)_{t \geq 0}$ is not a martingale with respect to $\mathcal{F}_t$.
Typically, the two properties, which you mentioned, hold true if $\mathcal{F}_t$ is the canonical filtration enlarged by some independent events. For instance, if, say, $U$ is a random variable which is independent from $(B_t)_{t \geq 0}$, then
$$\mathcal{F}_t := \sigma(U, B_r; r \leq t)$$
is strictly larger than the canonical filtration, and both properties (independence and martingale property) hold true.
