Since 1.) is already solved here, I will answer 2.):
$ \begin{eqnarray} X^T C-2AX &=& B &(1)\\ C^T X - 2X^T A^T &=& B^T &(2)\\ \end{eqnarray} $ Now add $(1)$ and $(2)$: $ \begin{eqnarray} X^T\underbrace{\left(C-2A^T\right)}_{D}+\underbrace{\left(C^T-2A\right)}_{D^T}X&=&B+B^T \tag{1+2}.\\ \end{eqnarray} $ Now let's multiply by $D^{-1}$ from the right: $ X^T + D^T X D^{-1} = \left(B+B^T\right)D^{-1} $ and use the following (to me known as superoperator formalism) representation of the problem: $ \text{vec}(AXB) = (B^T \otimes A) \text{vec}(X). $ (see here for a definition of $\text{vec}(X)$...). We get: $ \hat{T}\text{vec}(X) + \left((D^{-1})^T\otimes D \right) \text{vec}(X)=\text{vec}\left((B+B^T)D^{-1}\right), $ where $\hat{T}$ is the superoperator representation of the transposition operation (essentially a permutation matrix, that is not representable as product $A\otimes B$).
We finally get: $ \text{vec}(X)= \left[\hat{T} + \left((D^{-1})^T\otimes D \right) \right]^{-1}\text{vec}\left((B+B^T)D^{-1}\right) $