Let $\{X_k\}_{k\in [0,T]}$ be a Brownian motion - adapted to the sigma-algebra $\sigma_k$. Let $s<t$.
Why is $\Bbb{E}[(X_t-X_s)^2|\sigma_s]=t-s$?
I can see that $\Bbb{E}[X_t-X_s|\sigma_s]=0$ since $X_t-X_s$ is independent of $\sigma_s$ so that we have: $$\Bbb{E}[X_t-X_s|\sigma_s]=\Bbb{E}[X_t-X_s]\sim \mathcal{N}(0,t-s),$$ and with mean $0$ we obtain $\Bbb{E}[X_t-X_s|\sigma_s]=0.$
Not sure where to start for this one though.
With the same argument you can conclude: $$\Bbb{E}[(X_t-X_s)^2|\sigma_s] = \Bbb{E}[(X_t-X_s)^2] = \text{Var}(X_t - X_s) + \left(\Bbb{E}[X_t-X_s]\right)^2 = t-s$$ because $$X_t - X_s \sim \mathcal{N}(0,t-s)$$