Given joint density function of $(X,Y)$: $$f_{X,Y}(x,y) = \frac{x^3}{2} e^{-x(y+1)} \ \text{ for: } \ x,y > 0$$
Find $\mathbb{E} (Y|X)$, $\mathbb{E} (Y^2|X^2)$ and $\mathbb{P} (Y>1|X^3 + 1)$.
For $\mathbb{E} (Y|X)$:
$$f_X(x) = \int_y f_{X,Y}(x,y) \ dy = \int_0^{\infty} \frac{x^3}{2} e^{-x(y+1)} \ dy = \frac{x^2 e^{-x}}{2} \ \text{ when: } \ x>0 \ \text{ and } 0 \text{ otherwise. }$$
So we have that:
$$\mathbb{E}(Y|X) = \int_0^{\infty} \frac{y \cdot f_{X,Y}(x,y)}{f_X(x)} \ dy = \int_0^{\infty} \frac{y \cdot \frac{x^3}{2} e^{-x(y+1)}}{\frac{x^2 e^{-x}}{2}} \ dy = \int_0^{\infty} yxe^{-xy} \ dy = \frac{1}{x}$$
For $\mathbb{E} (Y^2|X^2)$:
As @GérardLetac said, since we know that $X > 0$, when we have $X^2$ that implies that we have $X$.
Therefore we have that
$$h(X) = \mathbb{E}(Y^2|X) \implies g(X^2) = h( \sqrt{X^2}) = \mathbb{E}(Y^2|X^2)$$
So we need to calculate only $\mathbb{E}(Y^2|X)$:
$$\mathbb{E}(Y^2|X) = \int_0^{\infty} \frac{y^2 \cdot f_{X,Y}(x,y)}{f_X(x)} \ dy = \int_0^{\infty} \frac{y^2 \cdot \frac{x^3}{2} e^{-x(y+1)}}{\frac{x^2 e^{-x}}{2}} \ dy = \int_0^{\infty} y^2xe^{-xy} \ dy = \frac{2}{x^2}$$
And we have that: $\mathbb{E}(Y^2|X^2) = \frac{2}{x^2}$
For $\mathbb{P} (Y>1|X^3 + 1)$:
As @GérardLetac said:
$$ h(X) = \mathbb{P}(Y > 1| X) \implies \mathbb{P}(Y > 1| X^3 + 1) = g(X) = h((X-1)^{\frac{1}{3}}) $$
Therefore we need to calculate only $\mathbb{P}(Y > 1| X)$ and put $(x-1)^{\frac{1}{3}}$ in place of $x$: $$\mathbb{P}(Y > 1| X) = \int_1^{\infty} \frac{f_{X,Y}(x,y)}{f_X(x)} dy = \int_1^{\infty} \frac{\frac{x^3}{2} e^{-x(y+1)}}{\frac{x^2 e^{-x}}{2}} dy = \int_1^{\infty} xe^{-xy} dy = e^{-x}$$
So we have that $\mathbb{P}(Y > 1| X^3 + 1) = e^{-\sqrt[3]{x-1}}$
Generate $X\sim\text{Gamma}(k=3, \theta=1)$, and $Y\vert X=x\sim \text{Exponential}(\lambda=x).$
You can check that this gives the right joint density: $$f(x,y)=f_x(x)f_{y|x}(y\vert x) \propto (x^{3-1}e^{-x})(xe^{-yx}) = x^3e^{-x(y+1)}.$$ Using well-known properties of the Exponential distribution, we have $$\mathbb{E}\left[Y\vert X=x\right]= \frac{1}{x}.$$
Using your/@GérardLetac's observation that $x\mapsto x^2$ and $x\mapsto x^3+1$ are 1-1 functions for $x>0$, we similarly compute
$$ \mathbb{E}\left[Y^2\vert X=x\right]= \frac{2!}{x^2}=\frac{2}{x^2},$$
and
$$P(Y>1\vert X=x)= e^{-x},$$
hence $$P(Y>1\vert X^3+1=u)=e^{-\sqrt[3]{u-1}}.$$