The book I'm reading does the following two computations which I don't follow:
Let $(B_t)_T$ be standard brownian motion. Consider the functions $$\phi_1(t,\omega) = \sum_{k=0}^{n-1}B_{t_k}\mathbb{1}_{(t_k,t_{k+1})} \\ \phi_2(t,\omega) = \sum_{k=0}^{n-1}B_{t_{k+1}}\mathbb{1}_{(t_k,t_{k+1})}$$
Then: $$E\bigg[\int^T_0\phi_1(t,\omega)dB_t(\omega)\bigg] = \sum_{k=0}^{n-1}E[B_{t_{k}}(B_{t_{k+1}}-B_{t_k})]$$ I understand the computation to this point, but then the book says this is $0$ by independence of increments. How do I see this? Do I know $B_{t_k}$ is independent of $B_{t_{k+1}}-B_{t_k}$? If so, then it's clear why this is zero.
Similarly:
$$E\bigg[\int^T_0\phi_2(t,\omega)dB_t(\omega)\bigg] \\ = \sum_{k=0}^{n-1}E[B_{t_{k+1}}(B_{t_{k+1}}-B_{t_k})] \\ = \sum_{k=0}^{n-1}E[(B_{t_{k+1}}-B_{t_k})^2]$$
How did they get from the second line to the third line? Why doesn't the independence of increments argument work here?
Yes you do know $B_{t_k}$ and $B_{t_{k+1}}-B_{t_k}$ are independent, as a basic property of a standard Wiener process
For your later question, while $B_{t_k}$ and $B_{t_{k+1}}-B_{t_k}$ are independent, $B_{t_{k+1}}$ and $B_{t_{k+1}}-B_{t_k}$ are not. So, also using $E[B_{t_{k+1}}]=E[B_{t_{k}}]=0$ to produce $E[B_{t_k}(B_{t_{k+1}}-B_{t_k})]=0$, you have
$$E[B_{t_{k+1}}(B_{t_{k+1}}-B_{t_k})]$$ $$=E[(B_{t_{k+1}}-B_{t_k}+B_{t_k})(B_{t_{k+1}}-B_{t_k})]$$ $$=E[(B_{t_{k+1}}-B_{t_k})^2]+E[B_{t_k}(B_{t_{k+1}}-B_{t_k})]$$ $$=E[(B_{t_{k+1}}-B_{t_k})^2]+0$$