Without the knowledge of independence, we cant solve this problem. So assuming $\displaystyle {X_i}$ are all independent.
$\displaystyle Y = \sum_{i=0}^{i=n} a_i X_{i} \implies \mathbb{E}(Y) = \sum_{i=0}^{i=n} a_i \mathbb{E}(X_{i})$and $\displaystyle \text{Var}(Y) = \sum_{i=0}^{i=n} a_i ^2 \text{Var}(X_{i})$
Without independence we could still solve it if we knew that $\displaystyle ({X_1},...,{X_n})$ was a Gaussian vector (which is not implied by the $\displaystyle {X_i}$ being all Gaussians). In that case, the mean of $\displaystyle Y$ would be the same, but its variance would be different (you should take covariances into account).