We have (X,Y) random vector with density $$ g_{(X,Y)}(x,y)= \begin{cases} \frac{y^3}{2}e^{-y(x+1)} & x, y \geq 0 \\ 0 & \text{otherwise} \end{cases} $$
Find $\mathbb{E}(X|Y), \mathbb{E}(X^2|Y^2), \mathbb{P}(X > 1|Y^3 + 1)$
Now, the first one was relatively easy:
I have found $g_Y(y)$ by performing: $$ g_Y(y)=\int_0^{\infty}\frac{y^3}{2}e^{-y(x+1)}dx =\frac{y^2}{2}e^{-y} $$ Then $$ \mathbb{E}(X|Y) = \int_0^{\infty}x \frac{\frac{y^3}{2}e^{-y(x+1)} }{\frac{y^2}{2}e^{-y}}dx = \int_0^{\infty}xye^{-xy}dx = \frac{1}{y} $$
Finally the answer is $\frac{1}{Y}$.
When looking for $\mathbb{E}(X^2|Y^2)$ I though that we need $g_Z(z)$ where $Z = Y^2$.
$$ F_Z(t) = \mathbb{P}(Z \leq t) = \mathbb{P}(Y \leq \sqrt{t}) = F_Y(\sqrt{t}) $$ Taking derivative w.r.t. t we get $$ g_Z(t)=\frac{1}{2\sqrt{t}}\frac{t}{2}e^{\frac{t}{2}} = \frac{\sqrt{t}}{4}e^{\frac{t}{2}} $$ How to proceed from there? I wanted to do something like $$ \mathbb{E}(X^2|Y^2) = \int_0^\infty x^2 \frac{g_{(X^2, Y^2)}(x,y)}{g_{Y^2}(y)}dx $$ but I have no idea how to get $g_{(X^2, Y^2)}(x,y)$.
What about $\mathbb{P}(X > 1|Y^3 + 1)$? Thank you for your time.
Your answer for the first part is mainly correct, save that the conditional expectation is a function of $Y$, rather than $y$.$$\mathsf E(X\mid Y)=\dfrac{\int_\Bbb R x~g_{\small X,Y}(x,Y)~\mathrm d x}{\int_\Bbb R g_{\small X,Y}(x,Y)~\mathrm d x}=\dfrac 1Y$$
For the rest, note, since $Y>0$ is certain, then when given either $Y^2$, or $Y^3+1$, you are also given $Y$. $$\mathsf E(X^2\mid Y^2)=\mathsf E(X^2\mid Y)=\dfrac{\int_\Bbb R x^2~g_{\small X,Y}(x,Y)~\mathrm d x}{\int_\Bbb R g_{\small X,Y}(x,Y)~\mathrm d x}$$
Though, express the answer in terms of $Y^2$
Likewise $$\mathsf P(X>1\mid Y^3+1)=\mathsf P(X>1\mid Y)=\dfrac{\int_1^\infty g_{\small X,Y}(x,Y)~\mathrm d x}{\int_0^\infty g_{\small X,Y}(x,Y)~\mathrm d x}$$
And express the final answer in terms of $Y^3+1$