This is about expectations of brownian motion and how they are connected to normal distribution.
I know that $B(t)$ is normal with mean $t$ and variance $t$ and that $E(B(t))=0$ if $(B(t))$ is a standard brownian motion since $B(t)$ has mean $0$ and variance $t$), but why is $E(B(t)^2)=t$?
Another example is why is $E(\exp(\sigma B(t)))=\exp(\sigma^2t/2)$, for $\sigma >0 $?
So, what would be $E(B(t)^3)$, for example? Or, if you have another useful example to help me understand that would be great.
Since $B(t)$ is normal $(0,t)$ (but not normal $(t,t)$, please be more careful...), one has $E[B(t)]=0$ and $\mathrm{var}(B(t))=t$ by definition of the parameters of a normal distribution. Likewise, $E[B(t)^3]=0$ and actually, $E[B(t)^{2n+1}]=0$ for every nonnegative integer $n$, since every centered normal distribution is even.
Finally, again by definition, $$E[\mathrm e^{\sigma B(t)-\sigma^2t/2}]=\mathrm e^{-\sigma^2t/2}E[\mathrm e^{\sigma B(t)}],$$ and $B(t)=\sqrt{t}X$ where $X$ is standard normal hence $$ E[\mathrm e^{\sigma B(t)}]=\int_\mathbb R\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}\mathrm e^{\sigma\sqrt{t}x}\mathrm dx=\int_\mathbb R\frac1{\sqrt{2\pi}}\mathrm e^{-(x-\sigma\sqrt t)^2/2}\mathrm e^{\sigma^2t/2}\mathrm dx=\mathrm e^{\sigma^2t/2}. $$ This yields $$ E[\mathrm e^{\sigma B(t)-\sigma^2t/2}]=1. $$ Actually, for every $\sigma$, the process $(M_t)_{t\geqslant0}$ defined by $M_t=\mathrm e^{\sigma B(t)-\sigma^2t/2}$ is a well-known martingale, called an exponential martingale.