Hello To prove that $(]1,+\infty[,d)$ where $d(x,y)=\left|\frac{1}{\ln(x)}-\frac{1}{\ln(y)}\right|$ is not complet i consider this sequence $(x_n)=\exp(n)$

it is a Cauchy sequence because

$$\lim_{p,q\to+\infty} \left|\frac{1}{\ln(\exp(p))}-\frac{1}{\ln(\exp(q))}\right|=\lim_{p,q\to+\infty}\left|\frac{1}{p}-\frac{1}{q}\right|=0$$

but it is not convergent.
if we suppose that it converge to $\ell>1$ then


\begin{align*}
\lim_{n\to+\infty} d(x_n,\ell)=0&\Longleftrightarrow
\lim_{n\to+\infty} \left|\frac{1}{\ln(x_n)}-\frac{1}{\ln(\ell)}\right|=0

proving that they're jointly gaussian should be a matter of calculation, but you can also skip that if you hide the calculation in theorems. Since X and Y are independent, (X,Y) is jointly gaussian and then
$$ \begin{pmatrix}
X+Y \\
X-Y
\end{pmatrix}= \begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix} \begin{pmatrix}
X \\
Y
\end{pmatrix}$$ is, as a (nonsingular) linear transformation of a gaussian variable, also gaussian with covariance matrix $\begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix} \begin{pmatrix}