The way you are presenting your question is a bit confusing. In particular, you're giving two different definitions for $X_t$. Here I'm going to use
\begin{align*}
Y_t &= \exp(\alpha W_t)\\
X_t &= X_0\exp((\mu-\frac{\sigma^2}{2})t + \sigma W_t)
\end{align*}
Note that in your question, I think there's a typo with the brackets in the definition of $X_t$. Also, everything I do here is under the assumption that $\alpha$, $\mu$ and $\sigma$ are in $\mathbb{R}$. I'm not sure how it applies when they are in $\mathbb{C}$, but from what I see, I don't think you need that now.
Now using Ito's lemma (https://en.wikipedia.org/wiki/It%C3%B4's_lemma) on $f(t,x)=e^{\alpha x}$ and $g(t,x) = e^{(\mu-\frac{\sigma^2}{2})t + \sigma x}$, you have that
\begin{align}
dY_t = \alpha Y_t dW_t + \frac{\alpha^2}{2} Y_t dt\\
dX_t = \mu X_t dt + \sigma X_t dW_t.
\end{align}
I'm unsure where you get the SDE you give for $dX_t$. Maybe there's something missing in your question.
Finally, for the expectation, what you have is the moment generating function of $V \sim N(\mu,\sigma^2)$ (see https://en.wikipedia.org/wiki/Normal_distribution#Moments). You can use this result to obtain $E[X_t]=X_0 e^{\mu t}$ as follows:
\begin{align*}
E[X_t] &= E[X_0 \exp((\mu-\frac{\sigma^2}{2})t + \sigma W_t)]\\
&= X_0 \exp((\mu-\frac{\sigma^2}{2})t) E[\exp(\sigma W_t)]\\
&= X_0 \exp((\mu-\frac{\sigma^2}{2})t) \exp(\frac{\sigma^2 t}{2})\\
&= X_0 e^{\mu t}.
\end{align*}
The third equality follows from $W_t \sim N(0,t)$. Analogously, $E[Y_t] = e^{\frac{\alpha^2 t}{2}}$.
