Calculate $\mathbb E(Y^2\mid X)$

The important fact to use is that $X$ and $Y$ are bivariate normal. Multivariate normal distributions have nice conditional distributions.

In particular, if you instead let $X$ and $Z$ be i.i.d. standard normal, then you can check that if you re-define $Y$ as $$\frac{1}{\sqrt{5}} Y := \rho X + \sqrt{1 - \rho^2} \cdot Z$$ with $\rho = -\frac{2}{\sqrt{5}}$, then $X$ and $Y$ are bivariate normal with the expectations/variances/covariance given in the question. (Check this.)

With this formulation, the conditional distribution of $Y$ given $X$ is easy to obtain. Given $X=x$, we have $$(Y \mid X=x) \overset{d}{=} \sqrt{5} \rho x + \sqrt{1 - \rho^2} \cdot Z \sim N(\sqrt{5} \rho x, 1-\rho^2)$$ so $\text{Var}(Y \mid X) = 1-\rho^2$ and $E[Y \mid X] = \sqrt{5} \rho X$, from which you can compute $E[Y^2 \mid X]$.