Let $Z = X_1 + X_2$ and let $W = X_1 - X_2$. From the question, both $Z$ and $W$ are standard normals. It is straight forward to show that
$$X_1 = \dfrac{Z+W}{2}$$ $$X_2 = \dfrac{Z-W}{2}$$
The mean of $X_1$ and $X_2$ is 0 (why?)
Because $Z$ and $W$ are standard normal and are assumed independent, then
$$\operatorname{Var}(X_1) = 0.25(\operatorname{Var}(Z) + \operatorname{Var}(W) + \operatorname{Cov}(Z,W)) = 0.25(1 + 1 + 0)= 0.5$$
A similar argument can be made for $X_2$.
We know $X_1$ and $X_2$ must be normal through the properties you've listed (that is, because $Z$ and $W$ are normal and $X_1$ and $X_2$ can be obtained from $Z$ and $W$ through addition/subtraction).
So we know three things:
- $X_1$ and $X_2$ are normal
- The mean of $X_1$ and $X_2$ is 0
- The variance $X_1$ and $X_2$ is 0.5
Can you finish from here?
Alternatively, consider $Z$ and $W$ as defined above and that they are standard normal. By definition of multivariate standard normals, the joint distribution of $Z$ and $W$ is multivariate standard normal. One can show that $x = Ay$ where $x$ is defined as above, $y = (Z,W)$ and
$$ A = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{bmatrix} $$
By definition, $x$ is multivariable normal with mean $(0,0)$, and with covariance matrix $\Sigma = AA^T$. If I remember correctly, $A$ is called a Cholesky Factor of $\Sigma$.