We select a number $X$ at random from $\{-3 , -1 , 0 , 2 , 5 \}$. If $X\le 0$ then we choose $Y$ randomly from $[X-2 , X+3]$ otherwise from $[-X , X]$ interval. Calculate value of $\mathbb{E}[Y]$.
My try: I think we should use iterated law of expectation. Namely, $\mathbb{E}[Y] = \mathbb{E}[\mathbb{E}[Y|X]]$ and then $\mathbb{E}[\mathbb{E}[Y|X]] = \sum \mathbb{E}[Y|x_k]P_X(x_k)$ but I don't know how to compute $\mathbb{E}[Y|x_k]$ and also this way seems too long.
Yes, use the law of iterated expectation. Condition on the partition $\{\{X\le 0\},\{X>0\}\}$ to get
$$\mathbb E[Y]=\mathbb E[Y\mid X\le 0]\mathbb P\{X\le 0\}+\mathbb E[Y\mid X>0]\mathbb P\{X>0\}$$
When $X\le 0$, it is given that $Y$ has a uniform distribution on $[X-2,X+3]$; otherwise $Y$ is uniform on $[-X,X]$. This information gives you
\begin{align} \mathbb E[Y\mid X\le 0]&=\sum_{j\le 0} \mathbb E[Y\mid X=j]\mathbb P\{X=j\mid X\le 0\} \\&=\frac1{\mathbb P\{X\le 0\}}\sum_{j\in\{-3,-1,0\}} \mathbb E[Y\mid X=j]\mathbb P\{X=j\} \\&=\frac1{\mathbb P\{X\le 0\}}\sum_{j\in\{-3,-1,0\}} \left(\frac{j-2+j+3}{2}\right)\cdot\frac15 \\&=\cdots \end{align}
Since $\mathbb E[Y\mid X=j]=\frac{-j+j}{2}=0$ for $j\in\{2,5\}$, you have $\mathbb E[Y\mid X>0]=0$ by a similar argument.