For the
skewness 3, notice that the graph is symmetrical about the mean and so we would expect a skewness of 0. Just to check it out
\begin{align*}
\text{Numerator = } & E[X^3] - 3 \mu E[X^2] + 2\mu^3\\
& = \int_{-1}^1 x^3 \cdot \frac{3}{4} \cdot (1-x^2) dx - 3 E[X^2] \cdot 0 + 0^3 \\
& = \int_{-1}^1 \cdot \frac{3}{4} \cdot (x^3-x^5) dx\\
& = \frac{3}{4} \cdot (x^4/4-x^6/6) \big |_{-1}^1\\
& = 0
\end{align*}
as expected without having to actually complete the calculation by dividing by the cube of the standard deviation.