I believe:
$$P(X\ge 0) = \int_0^\infty f(x)dx =\int_0^2 \dfrac{x^2}{3}dx = \dfrac{8}{9}$$
$$E[Y] = E[X^2] = \int_{-\infty}^\infty x^2f(x)dx = \int_{-1}^2 \dfrac{x^4}{3}dx = \left. \dfrac{x^5}{15}\right|_{-1}^2 = \dfrac{11}{5}$$
$$V[Y] = E[Y^2] - (E[Y])^2 = E[X^4] - (E[Y])^2 = \int_{-1}^2 \dfrac{x^6}{3}dx-\dfrac{121}{25} = \dfrac{43}{7}-\dfrac{121}{25} = \dfrac{228}{175}$$
I could be wrong, but that is how I would do it.