Suppose that $X$ and $Y$ are distributed as independent Standard Normals. Find the distribution of $(X-Y, X+Y)$.
Isn't the case for $X-Y$ elementary? Since they are both standard normals, this should equate to $0$ for all components concerning this portion of the vector?
Additionally, when it says "find the distribution", would this just mean me finding the mean, variance and covariance of each component?
If $X, Y$ are independent and have the standard normal distribution, then $X-Y$ has normal distribution of mean $E(X-Y)=E(X)-E(Y)=0$ and dispersion $\sigma^2(X-Y)= \sigma^2(X)+\sigma^2(Y)=2$. Analogously, $X+Y\sim N(E(X)+E(Y), \sigma^2(X)+\sigma^2(Y))$, i.e. $X+Y\sim N(0,\sigma^2=2)$.
In order to get the joint distribution of $(X-Y, X+Y)$ we have to compute its mean vector and the matrix of covariance, $\Sigma$. The mean vector $m=(0,0)^T$. To get $\Sigma_{12}=\Sigma_{21}$ we need the covariance: $cov(X-Y, X+Y)=E((X-Y)(X+Y))-E(X-Y)E(X+Y)=E(X^2-Y^2)-[(E(X))^2-(E(Y))^2]=E(X^2)-(E(X))^2-[E(Y^2)-(E(Y))^2]=\sigma^2(X)-\sigma^2(Y)=0$. Hence the two random variables are uncorelated, and thus $\Sigma=\left[\begin{array}{cc} \sigma^2(X-Y)&0\\ 0&\sigma^2(X+Y)\end{array}\right]=\left[\begin{array}{cc}2&0\\0&2\end{array}\right]$. Now you can write down the joint density distribution function.