= Solution
Write the <expected value> as $\mu^T=(\mu_1^T,\mu_2^T)$ and assume the relevant <covariance matrix> $V_{22}$ is invertible. Remove the linear predictor from the first block by defining
$$
Y=X_1-\mu_1-V_{12}V_{22}^{-1}(X_2-\mu_2).
$$
A <linear image of a multivariate normal vector> is again <multivariate normal>. Direct <covariance> calculation gives
$$
\operatorname{Cov}(Y,X_2)=V_{12}-V_{12}V_{22}^{-1}V_{22}=0.
$$
Thus $Y$ and $X_2$ are <independent>: <uncorrelated jointly Gaussian variables are independent>. Expanding the <covariance matrix> of $Y$, using $V_{21}=V_{12}^T$, gives
$$
\begin{aligned}
\operatorname{Cov}(Y)
&=V_{11}-V_{12}V_{22}^{-1}V_{21}-V_{12}V_{22}^{-1}V_{21}
+V_{12}V_{22}^{-1}V_{22}V_{22}^{-1}V_{21}\\
&=V_{11}-V_{12}V_{22}^{-1}V_{21}.
\end{aligned}
$$
Conditioning on $X_2=x_2$ translates $Y$ by the fixed vector $\mu_1+V_{12}V_{22}^{-1}(x_2-\mu_2)$ and leaves its <covariance matrix> unchanged. Consequently the <conditional multivariate normal distribution> is
$$
X_1\mid X_2=x_2\sim N\!\left(\mu_1+V_{12}V_{22}^{-1}(x_2-\mu_2),\ \boxed{V_{11}-V_{12}V_{22}^{-1}V_{21}}\right).
$$
The <conditional covariance> is the <Schur complement> and does not depend on $x_2$.
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