Let's assume you mean $X\sim\mathcal{N}(\mu_X, \sigma_X)$ and that $y$ is a constant chosen independently of observing $X$. Reduce the problem to finding the conditional expectation of $Z = X-y$ conditional on $Z \ge 0$: adding $y$ to that value gives the desired answer. (Whether $y$ is positive is immaterial.)
The governing property of conditional probability is the multiplicative relationship
$$\Pr(Z\in\mathcal{A}\,|\,Z \ge 0)\Pr(Z \ge 0) = \Pr(Z\in\mathcal{A\cap[0,\infty)})$$
for all measurable sets $\mathcal{A}$. In particular, letting $\mathcal{A}=(z,\infty)$ for some $z\ge 0$, solve for the conditional probability:
$$\Pr(Z \gt z\,|\,Z \ge 0) = \frac{\Pr(Z \gt z)}{\Pr(Z \ge 0)}.$$
The left hand side is the conditional survival function while the numerator and denominator on the right are both in terms of the survival function of $Z$ itself. Write $\Phi(z; \mu, \sigma)$ for the Normal distribution function with mean $\mu$ and standard deviation $\sigma$. Its complement $1-\Phi$ is the survival function. Because $Z$ obviously is Normal with mean $\mu_X-y$ and standard deviation $\sigma_X$, the survival function of the positive part of $Z$, $Z^{+}$, is
$$S_{Z^{+}}(z) = \frac{1-\Phi(z; \mu_X-y, \sigma_X)}{1 - \Phi(0; \mu_X-y, \sigma_X)}$$
for $z \ge 0$. Its integral gives the conditional expectation. Add back $y$ to give the answer
$$y + \frac{1}{1 - \Phi(0; \mu_X-y, \sigma_X)}\int_0^\infty \left(1 - \Phi(z; \mu_X-y, \sigma_X)\right)dz.$$
As the integral of a complementary error function, it has no simpler expression in general.
You have the correct formulas, but I leave it to you to check whether you've applied them correctly.
As for the distribution of $(2X−Z,3Y+Z)$, viewed as a 2 element column vector.
Consider$(X.Y,Z)$ as a 3 element column vector. You need to determine the matrix $A$ such that $A*(X,Y,Z) = (2X−Z,3Y+Z)$ . Hint: what dimensions must $A$ have to transform a 3 by 1 vector into a 2 by 1 vector? Then use the result $\text{Cov} (A*(X,Y,Z)) = A* \text{Cov}(X,Y,Z)*A^T$ combined with the trivial calculation of the mean, and your knowledge of the type of distribution which a linear transformation of a Multivariate Gaussian has.
Best Answer
What is meant is simply that for Gaussian variables: dependency = linear dependency. In other words, all the information on the dependency between two Gaussians is in their covariance or correlation.
In turn, if you know the covariance matrix, then you have the variances and correlations between each element $X_i$ and the others: $\rho(X_j, X_j), j \neq i$.
Last, if you have centered gaussians and you know their variances and correlations, you can write them using independent Gaussians which will give you the expression of the conditional value of one given the others.
Let's take two standard guassians for example $(X_1, X_2)$ with a correlation $\rho$. If you want the distribution of $X_2 | X_1$ for example, you have to:
write $X_2 = \rho X_1 + \sqrt{1 - \rho^2} X_0$ , with $X_0$ a standard gaussian independant from $X_1$.
conclude that $X_2 | X_1$ is gaussian with mean $\rho X_1$ and variance $1 - \rho^2$.
This extends to an inversion of Cholesky's decomposition (a.k.a. square root matrix of the covariance matrix) in higher dimensions.