No, your answers are wrong.
When events $A$ and $B$ are independent, then $\mathbb{P}(A \cap B) = \mathbb{P}(A) \mathbb{P}(B)$, and $\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B) = 1 -(1-\mathbb{P}(A))(1-\mathbb{P}(B)))$.
Let $T_X$ denote the event that child $X$ will get the ticket.
For the first part, none of the kids get the tickets is logically equivalent to (X does not get a ticket) and (Y does not get a ticket) and (Z does not get the ticket). Since these events are independent, $\mathbb{P}( \lnot T_X \land \lnot T_Y \land \lnot T_Z) = \mathbb{P}( \lnot T_X ) \mathbb{P}( \lnot T_Y ) \mathbb{P}( \lnot T_Z) = (1 - \mathbb{P}( T_X))(1 - \mathbb{P}( T_Y))(1 - \mathbb{P}( T_Z))$. So $\mathbb{P}( \lnot T_X \land \lnot T_Y \land \lnot T_Z) = (1-0.4)(1-0.3)(1-0.2) = 0.336$.
For the second question, the logical expression for Only one will get the ticket is
$(T_X \land \lnot T_Y \land \lnot T_Z) \lor (\lnot T_X \land T_Y \land \lnot T_Z) \lor (
\lnot T_X \land \lnot T_Y \land T_Z)$. I will leave to you to work it out, the answer should be $0.452$.
For the third question, it is easier to think of the logical opposite, which you have already answer in earlier question.
For the last part, the logical expression is $T_X \land T_Y \land T_Z$, so the probabilities are products $0.4 \times 0.3 \times 0.2 = 0.024$.
When $w_0=0$ they are not conditionally independent, and you have an example of Berkson's paradox. They are only conditionally independent when $w_0=1$, in which case everyone is weird.
I would say you may not have given an "explanation in words". Perhaps you could say something like
"When $w_0=0$ everyone with the weird symptom has at least one of the diseases. So anyone with the weird symptom but not $D_1$ must have $D_2$ while only a proportion $p_2$ of those with the weird symptom and $D_1$ are also expected to have $D_2$, and thus $D_1$ and $D_2$ are not conditionally independent given the weird symptom."
Best Answer
I will just do the first part, because the other parts are similar.
You want to find the probability that $J_3$ votes guilty given the info that $J_1$ and $J_2$ votes guilty, and for this, we first need the probability that $J_1$ and $J_2$ even votes guilty in the first place.
Let $J_{ig}$ denote the event that judge $i$ votes guilty.
\begin{align} P(J_{1g}, J_{2g})&=P(J_{1g}, J_{2g}|\text{defendant is guilty})P(\text{defendant is guilty})\\ &+P(J_{1g}, J_{2g}|\text{defendant is innocent})P(\text{defendant is innocent}) \\ &=\left(\frac{7}{10}\right)^2\left(\frac{7}{10}\right)+\left(\frac{3}{10}\right)^2\left(\frac{2}{10}\right)\\ &=\frac{71}{200} \end{align}
Moving on to the required probability: \begin{align} P(J_{3g}|J_{1g}, J_{2g})&=\frac{P(J_{3g}, J_{1g}, J_{2g}|\text{def guilty})P(\text{def guilty})+P(J_{3g}, J_{1g}, J_{2g}|\text{def innocent})P(\text{def innocent})}{P(J_{1g}, J_{2g})}\\ &=\frac{\left(\frac{7}{10}\right)^3\left(\frac{7}{10}\right)+\left(\frac{2}{10}\right)^3\left(\frac{3}{10}\right)}{\frac{71}{200}}\\ &=\frac{97}{142} \end{align}