We have ${x_1, x_2}$ ~ ${N(0,1)}$ and it is not said that they are independent.
Show that then ${(x_1, x_2)}$ ~ ${N(a,b)}$ , where a - vector of mean values and b is covariance matrix.
I know that if ${x_1, x_2}$ independent we can get bivariate normal distribution, but I don't know how to deal with knowing nothing about dependence.
If $x_1$ and $x_2$ are gaussian then $(x_1,x_2)$ is not necessarily gaussian, so you will not be able to prove your statement.
The classic counterexample is $x_1\sim N(0,1)$ and $x_2=\varepsilon x_1$, where $\varepsilon$ is independent of $x_1$ and $\mathbb P(\varepsilon=1)=\mathbb P(\varepsilon=-1)=1/2$.
It is easy to see that in this setting, $x_2\sim N(0,1)$. But $(x_1,x_2)$ is not gaussian since $x_1+x_2=(1+\varepsilon)x_1$ is equal to zero with probability $1/2$, so it is not gaussian.
