Please check if my answers are correct - statistic, probability

See a short discussion of the solutions below. In some cases I get the same number as you, just rounded differently. If you need additional decimal precision or if some part is not clear to you don't hesitate to ask.

1. The correct number is ${3\choose 3}\cdot {10\choose 3}\cdot 3!=720$ (choose $3$ from the $3$ men, $3$ from the $10$ women and then choose a way to pair the chosen $3$ men with the chosen $3$ women).
2. The correct number is ${13\choose 2}+{13\choose 3}+{13\choose 4}+{13\choose 5}= 2366$ (there are ${13\choose 2}$ many configurations where $2$ people goe in the second car and $11$ in the first, then ${13\choose 3}$ configurations where $3$ people go in the second car and $10$ in the first, etc)
3. The answer depends on what is meant with "4 defective devices in a sample of 9", this could be understood as "exactly 4 from 9 are defective" or "atleast 4 from 9 are defective". In the first case the probability is $$\frac{{23\choose 4}\cdot {37\choose 5}}{{60\choose 9}}\approx 0.2611$$ ie $26.11\%$. In the second case you get:
   $$\sum_{k=4}^9\frac{{23\choose k}\cdot {37\choose 9-k}}{60\choose 9}\approx 0.477$$ ie $47.7\%$.
4. Let $I$ denote the event that you are infected, and $T$ the event that the test is positive. Then
   $$P(I) = 0.0038,\quad P(T\mid I )= 0.994,\quad P(\neg T\mid \neg I) =0.94$$ we want to find $P(I\mid T)$. Using Bayes' theorem: $$P(I\mid T) = \frac{P(T\mid I) P(I)}{P(T)}$$ the number $P(T)$ is still unknown, but here we use the formula $$P(T) = P(T\mid I) P(I) + P(T\mid \neg I) P(\neg I) = P(T\mid I) P(I) +(1-P(\neg T\mid I)) (1-P(I))$$ in numberes then: $$P(T) = 0.994\cdot 0.0038+(1-0.94)\cdot (1-0.0038) \approx 0.06355$$ and then $$P(I\mid T) = \frac{0.994\cdot 0.0038}{0.06355}\approx 0.0594$$ so about $5.94\%$.
5. Using the same labels we now read off: $$P(T\mid I) = 0.98, \quad P(\neg T\mid \neg I)= 0.934, \quad P(I\mid T) = 0.896$$we want to find  $P(I)$, we use the formula: $$P(I) = \frac{P(I\mid T) P(T)}{P(T\mid I)}$$ Here we still need to find $P(T)$. Note that$$P(T) = P(T\mid I ) P(I) + P(T\mid \neg I) P(\neg I) = P(T\mid I) P(I) + (1-P(\neg T\mid\neg I)(1-P(I))$$ so $$P(T) = P(I) \cdot (P(T\mid I) + P(\neg T\mid \neg I) -1) + 1-P(\neg T\mid \neg I)= 0.914\cdot P(I) +0.076$$
   We put this into our equation for $P(I)$ and get
   $$P(I) = \frac{0.896\cdot (0.914 \cdot P(I)+0.076)}{0.98}\approx 0.8357\cdot P(I) +0.0695 $$
   so $P(I) \approx \frac{0.0695}{1-0.8357}\approx 0.423$, ie approximately $42.3\%$.
6. The probability is $$\sum_{k=0}^3 {23\choose k}0.16^k\cdot (1-0.16)^{23-k} \approx 0.486$$ or approximately $48.6\%$.
7. The probability is $\frac{(924!/920!)\cdot 76}{(1000!/995!)}\approx 0.0556$ or about $5.56\%$.
8. You need to buy more than $23$ lots if you draw $23$ failures in a row. Since we are approxmating this by using replacement we get:$$(\frac{924}{1000})^{23}\approx 0.1624$$ as a probablity, which is $16.24\%$.
9. The expected winnings of one ticket is:
   $$1000\cdot \frac{3}{1000}+ 200\cdot \frac{10}{1000}+10\cdot \frac{63}{1000}+0\cdot \frac{924}{1000} = 5.63$$
   note that to find the expected "profit" we should subtract the cost of a ticket from the expected winnings. But since no such cost is given I assume the word "profit" was just poorly chosen.