You can use the result that a linear combination of normally distributed random variables is also normal, in this case X is bivariate normal.
\begin{equation} \textbf{X}_{p=2} = \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} \in N_{p=2}( \bf{ \mu }, \bf{ \Sigma } ) = N\left( \begin{bmatrix} \mu_1 =1 \\ \mu_2 =1 \end{bmatrix} , \begin{bmatrix} 2 & \sigma_{12} \\ \sigma_{21} & 2 \end{bmatrix} \right) \end{equation}\begin{equation} \textbf{X}_{p=2} = \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} \in N_{p=2}( \bf{ \mu_x }, \bf{ \Sigma_x } ) = N\left( \begin{bmatrix} \mu_1 =1 \\ \mu_2 =1 \end{bmatrix} , \begin{bmatrix} 2 & \sigma_{12} \\ \sigma_{21} & 2 \end{bmatrix} \right) \end{equation} Then \begin{equation} \bf{A}\bf{X} = \begin{bmatrix} 1 & 1 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} = \begin{bmatrix} X_1 + X_2 \\ X_1 -2X_2 \end{bmatrix} = \begin{bmatrix} U \\ V \end{bmatrix} \in N_{p=2} \left(\bf{A}\bf{X}, \bf{A}\bf{\Sigma}\bf{A}' \right) \end{equation}\begin{equation} \bf{A}\bf{X} = \begin{bmatrix} 1 & 1 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} = \begin{bmatrix} X_1 + X_2 \\ X_1 -2X_2 \end{bmatrix} = \begin{bmatrix} U \\ V \end{bmatrix} \in N_{p=2} \left(\bf{A}\bf{\mu_x}, \bf{A}\bf{\Sigma_x}\bf{A}' \right) \end{equation} Finally calculate $\rho = \sigma_{12}/(\sqrt{\sigma_{11}}\sqrt{\sigma_{22}})$ from the entries of $\bf{A}\bf{\Sigma}\bf{A}'$, which is the correlation between U and V. Note that $\sigma_{11},\sigma_{22}$ denotes the variances (just notation).