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 $0.4 \times 0.3 \times 0.2 = 0.024$.
Let $N$ be the random number of hospitalizations, so $N \in \{0, 1, 2\} \sim \operatorname{Binomial}(n = 2, p = 0.3)$. Let $L$ be the total loss of all hospitalizations. We are asked to find $$\operatorname{E}[N \mid L \le 1] = \sum_{k=0}^2 k \Pr[N = k \mid L \le 1],$$ by the definition of expectation. Now we must compute the conditional probabilities. $$\Pr[N = k \mid L \le 1] = \frac{\Pr[L \le 1 \mid N = k]\Pr[N = k]}{\Pr[L \le 1]},$$ by Bayes' theorem. We know that $$\Pr[N = k] = \binom{2}{k} (0.3)^k (0.7)^{2-k},$$ and: $$\begin{align*} \Pr[L \le 1 \mid N = 0] &= 1, \\ \Pr[L \le 1 \mid N = 1] &= 1, \\ \Pr[L \le 1 \mid N = 2] &= \int_{x=0}^1 \int_{y=0}^{1-x} \, dy \, dx = \frac{1}{2}. \end{align*}$$ Therefore, $$\Pr[L \le 1] = 1(0.7)^2 + 1(2)(0.3)(0.7) + \frac{1}{2}(0.3)^2 = \frac{191}{200},$$ and the rest of the calculation is straightforward: $$\operatorname{E}[N \mid L \le 1] = \frac{\Pr[L \le 1 \mid N = 1]\Pr[N = 1] + 2 \Pr[L \le 1 \mid N = 2]\Pr[N = 2]}{\Pr[L \le 1]}.$$
Mathematica code to compute the empirical expectation for $10^6$ simulations:
Mean[Last /@ Select[{Total[RandomVariate[UniformDistribution[{0, 1}], #]], #} & /@ RandomVariate[BinomialDistribution[2, 0.3], 10^6], First[#] <= 1 &]]
Best Answer
For at least one stock to be a total loss, is for it to not be the case that all stocks are not total losses.
If the probability of $X$ happening is $P(X)$ then the probability of $X$ not happening is $1-P(X)$.
So the probability of at least one stock being a total loss is $1 -$ the probability that all stocks are not total losses.
If the probability of $X$ is $P(X)$, then the probability of $X$ occurring $n$ times out of $n$ is $(P(X))^n$.
So the probability that all stocks are not total losses is $($the probability one stock is not a total losses$)^4$.
And repeating myself: If the probability of $X$ happening is $P(X)$ then the probability of $X$ not happening is $1-P(X)$.
So the probability that one stock is not a total loss is $1 -$ probability that a stock is a total loss.
And probability that a stock is a total lose is $0.4$.