For
skewness 3, note that in computing the variance above you also found that
\begin{equation*}
E[X^2] = \frac{11}{5}.
\end{equation*}
\begin{equation*}
E[X^3] = \int_{-1}^2 x^3 \cdot x^2/3 dx = \frac{x^6}{18} |_{-1}^2 = \frac{7}{2}
\end{equation*}
and so
\begin{equation*}
\gamma_1 = \frac{\frac{7}{2} - 3 \cdot \frac{5}{4} \cdot \frac{11}{5} + 2 \cdot (\frac{5}{4})^3}{\sqrt{\frac{51}{80}}^3}
\end{equation*}