Hi Kathy, sorry it took a while to get back to you about these. I'll give some thoughts.

Question 1) is an example of a problem that uses what is called "Pigeonhole Principle", which is  a more advanced concept. I won't go into to many details here, but the idea is to think of a worst-case scenario.

For this problem, suppose you are choosing socks (still one by one) to give to the Martian. Your goal is to give him socks as long as possible, trying to make sure there are NOT five socks of the same color. (You can see the colors of the socks.) Try to understand why (1) you can do this for a total of $12$ socks, but (2) no matter how you choose the socks, if you give him a $13$th sock there will be a match.

The key idea for Question 2) is using a "divisibility rule" for $11$. As an example of a divisibility rule, we can easily tell whether a number is divisible by $2$ using its last digit. The rule for $11$ is a little more complicated, using what is sometimes called the "alternating sum" of the digits. Let me know if you want the specifics, but this is something we'll cover in the spring session of the Math Challenge courses, so I won't spoil it right now :)

Question 3) is a fairly challenging one, so I'd recommend holding off on it for now. We'll do some simpler versions of this type of problem during this session of class. The main reason the problem is difficult is that there are many cases to deal with. For example, if the friends are $A$, $B$, and $C$ we could give $1$ book to $A$ and $B$ and $5$ books to $C$, or $2$ books to $A$, $4$ books to $B$, and $1$ book to $C$, etc. And this is not even deciding which book to give to each! There is a fancier way of solving this problem, but that way uses a method is usually not taught until our Math Challenge II classes (it is called the Principle of Inclusion-Exclusion). Also the answer isn't $21$ (I think you might have seen it wrong) is is much larger:  $1806$, so don't try to list out all the possibilities!

Question 4) is a good question that you should be able to revisit and get correct at the end of this course! To get started, this is a good example of a question to do the opposite (or complement). For the product to be even, one or both of the spinners must last on an even number (so we need two cases). Rather than doing the two cases, we can look at the opposite, which is the product is odd, so both spinners must land on a odd number. This gets you on your way to solving the question.

Question 5) is another really tough question, that needs a decent amount of algebra to solve. (I'd probably recommend not worrying about this one right now either.) The first step is factoring the equation, and the hint I'll give is that you want to start by noticing the equation has the form $$(\underline{\ \ \ \ \ })^2 + 3x(\underline{\ \ \ \ \ }) + 2x^2 = 0$$ so we can start by thinking of it as a quadratic with the underlined parts missing.

For Question 6) we'll need the idea of combinations, which we'll be talking about soon in class. As Steve as $6$ bars of candy and he's bringing two per day, we just need to decide which candy bars he brings the first day, which the second day, and which the third day. Since Steve is just dividing his candy bars he doesn't care about the order of the $2$ candy bars he brings each day, so we use combinations: $$\binom{6}{2}\times \binom{4}{2}\times \binom{2}{2}.$$ (We are multiplying because he needs to decide for the first AND the second AND the third days.)

Hope these thoughts help a little bit, and let us know if you have more questions.