The expected value of $X$ can be computed as follows:

\begin{eqnarray*}
E[X] & = & \int_{-\infty}^{\infty} xf(x)dx \\
 & = & \int_{0}^{\infty} x \frac{x^{\alpha -1} e^{-x/\beta}} {\Gamma (\alpha) \beta^{\alpha}} dx \\
 & = & \frac{\Gamma(\alpha+1)\beta^{\alpha+1}}{\Gamma(\alpha)\beta^{\alpha}} \int_{0}^{\infty} \frac{x^{(\alpha+1) -1} e^{-x/\beta}} {\Gamma (\alpha+1) \beta^{\alpha+1}} dx\\
 & = & \frac{\alpha\Gamma(\alpha)\beta^{\alpha+1}}{\Gamma(\alpha)\beta^{\alpha}}
\end{eqnarray*}

so $E[X] = \alpha\beta$. \\

Next, the m.g.f.is given by

\begin{eqnarray*}
E[e^{tX}] & = & \int_{0}^{\infty} e^{tx}
                \frac{x^{\alpha-1}e^{-x/\beta}}
                     {\Gamma(\alpha)\beta^{\alpha}}
                dx \\
& = & \frac{1}{\Gamma(\alpha)\beta^{\alpha}}
\int_{0}^{\infty} x^{\alpha-1} e^{tx - x/\beta} dx \\
& = & \frac{\Gamma(\alpha) \phi^{\alpha} }
            {\Gamma(\alpha) \beta^{\alpha}}
\int_{0}^{\infty} \frac{x^{(\alpha-1)} e^{-x/\phi}} {\Gamma (\alpha) \phi^{\alpha}}dx
\end{eqnarray*}


if we choose $\phi$ so that $\frac{-x}{\phi} = tx - x/\beta$ i.e. $\frac{-1}{\phi} = t - \frac{1}{\beta}$
i.e. $\phi = - \frac{1}{t-1/\beta} = \frac{\beta}{1 - \beta t}$
then we have

\begin{eqnarray*}
M(t) & = & \left(\frac{\phi}{\beta}\right)^{\alpha} \\
& = & \left(\frac{\beta / (1-\beta t)}{\beta}\right)^{\alpha} \\
& = & \frac{1}{(1-\beta t)^{\alpha} }
\end{eqnarray*}

or $M(t) = (1-\beta t)^{-\alpha}$.
It follows that

$$ M'(t) = (-\alpha) (1-\beta t)^{-\alpha-1} (-\beta) = \alpha\beta(1-\beta t)^{-\alpha-1}$$

so $ M'(0) = \alpha\beta $. Further,

\begin{eqnarray*}
M''(t) & = & \alpha\beta (-\alpha-1)(1-\beta t)^{-\alpha-2} (-\beta) \\
& = & \alpha\beta^2 (\alpha+1)(1-\beta t)^{-\alpha-2}
\end{eqnarray*}

\begin{eqnarray*}
E[X^2] & = & M''(0) \\
& = & \alpha\beta^2 (\alpha+1) \\
& = & \alpha^2 \beta^2 + \alpha \beta^2
\end{eqnarray*}

Hence,

\begin{eqnarray*}
V[X] & = & E[X]^2 - E[X]^2\\
& = & \alpha^2 \beta^2 + \alpha \beta^2 - (\alpha\beta)^2 \\
& = & \alpha \beta^2
\end{eqnarray*}