Q: Compute expected value from two dimensional normal distribution 
Random vector (X, Y) has the two-dimensional normal distribution
with the density $ f(x,y)=\frac{1}{2\pi\sqrt{2}}\exp\left\{
 -\frac{1}{8} \left[ 4x^2 +8x(y+3)+6(y+3)^2 \right] \right\} $
Compute $\mathbb{E} \left[ XY - 2X - 3Y^2 + 1 \right ]$
Find the covariance matrix of the random vector $(U, V) = (-2X+Y, X-Y)$

Solution:
$det \; C = 2$
$
f(x,y)=\frac{1}{2\pi\sqrt{2}}\exp\left\{ -\frac{1}{4} \left[ 2x^2 +4x(y+3)+3(y+3)^2 \right] \right\}
$
$
m = \begin{pmatrix}
  0\\ 
  -3
\end{pmatrix} = \begin{pmatrix}
  \mathbb{E} X\\ 
  \mathbb{E} Y
\end{pmatrix}
$
$
C = \begin{pmatrix}
  3 & 2\\ 
  2 & 2
\end{pmatrix} = \begin{pmatrix}
  Var X & Cov(X, Y)\\ 
  Cov(X, Y) & Var Y
\end{pmatrix}
$
$
\mathbb{E} \left[ XY - 2X - 3Y^2 + 1 \right ] = \mathbb{E}[XY] - 2\mathbb{E}X - 3\mathbb{E}[Y^2] + 1
$
$
Cov(X,Y)=\mathbb{E}[XY]-\mathbb{E}X\mathbb{E}Y \rightarrow \mathbb{E}[XY]=2
$
$
Var Y = \mathbb{E}(Y^2)-(\mathbb{E}Y)^2 \rightarrow (\mathbb{E}Y)^2 = 11
$
$
\mathbb{E} \left[ XY - 2X - 3Y^2 + 1 \right ] =2-2\cdot0-3\cdot11+1=-30
$
$
A \cdot \begin{pmatrix}
  X\\ 
  Y
\end{pmatrix} = \begin{pmatrix}
  -2X & Y\\ 
  X & -Y
\end{pmatrix} \rightarrow A = \begin{pmatrix}
  -2 & 1\\ 
  1 & -1
\end{pmatrix}
$
$
C_{UV} = A \cdot C \cdot A^T = \begin{pmatrix}
  6 & -2\\ 
  2 & 1
\end{pmatrix}
$
I am asking for verification of the solution and corrections if needed.

A: The term in the exponent of $f(x,y)$ can be written as
$$\begin{split}-\frac 18\left[4x^2+8x(y+3)+6(y+3)^2\right] &= -\frac 12 \left[x^2+2x(y+3)+\frac 32(y+3)^2\right]\\
&=-\frac 12\begin{pmatrix}x\\
y+3\end{pmatrix}^T\begin{pmatrix}1&1\\1&3/2\end{pmatrix}\begin{pmatrix}x\\y+3\end{pmatrix}\end{split}$$
Thus $\Sigma^{-1}=\begin{pmatrix}1&1\\1&3/2\end{pmatrix}$ and the covariance matrix is $\Sigma=\frac 1{3/2-1}\begin{pmatrix}3/2&-1\\-1&1\end{pmatrix}=\begin{pmatrix}3&-2\\-2&2\end{pmatrix}$
Besides this, your calculations look correct, so just redo the parts that need changing.
