One relatively easy systematic approach is to make a table for the combinations; here I’ve listed the shirts down the side and the ties across the top. I’ve put a $b$ in the cells corresponding to acceptable combinations that include blue, and a $y$ in the cells corresponding to other acceptable combinations.
$$\begin{array}{r|c}
&\text{white}&\text{blue}&\text{yellow}&\text{green/blue}\\ \hline
\text{blue}&b&&b&b\\ \hline
\text{green}&y&b&y&b\\ \hline
\text{white}&&b&y\\ \hline
\text{blue/green}&&b\\ \hline
\text{red/white}&y
\end{array}$$
As you can see, there are indeed $11$ acceptable combinations, but seven contain blue. Thus, if he is equally likely to pick any of the acceptable combinations, his probability of picking one that includes blue is $\frac7{11}$.
I agree with you on Event A; I'm not so sure about event C. And I disagree with your Event B.
For Event B: consider the case where Aji eats his sandwich second to last. You claim that there are six ways of doing this; I think there are only 4 ways: the meat must be put into the sandwich either first or second; then you have two choices of what else to do before eating the sandwich; the last thing you do is now forced. Explicitly (with m=putting meat, f=eating fruit, j=drinking juice, and s=eating sandwich) the only four things he can do are mfsj, mjsf, fmsj, and jmsf. I don't see six things.
Where is your counting error? It is true that you have three choices for the first position, but you don't have two choices for the second thing. You only have two choices for the second thing if the first things you did was putting the meat in; if you chose anything else for your first thing to do, then the second thing to do must be putting the meat in the sandwich, so you don't really have 2 choices for the second thing regardless of the third. Instead, you would have to break it up into cases: you have 2 ways of doing it if the meat happens first; and 2 ways if the meat happens second.
Here's a simpler way of doing Event B:
Since Aji must put meat in the sandwich before eating it, we know that, whethever order he does things, "put meat" goes before "eat sandwich". There are things that can happen before he puts meat in the sandwich, things that can happen after he puts meat in the sandwich but before he eats it, and things that can happen after he eats his sandwich.
So, relative to the putting-meat-and-eating-his-sandwich, there are three temporal locations for "eat his fruit" to go into:
<location 1> put meat <location 2> eat sandwich <location 3>
Once we chronologically locate the fruit relative to the sandwich-eating process, what about the juice? There are now four temporal locations for the juice: any of the two temporal locations not used by the fruit; and the same temporal location as the fruit, but either before the fruit, or after the fruit. For instance, if the fruit was eaten between putting the meat and eating the sandwich, then the juice can be had before putting the meat, or after the meat but before the fruit; or after the fruit but before the sandwich; or after the sandwich.
This gives $3\times 4 = 12$ different ways of doing Event B.
As for Event C: note that the problem states that he picks up the fishing rod and the icebox after tying the vote. That means that the first thing Aji does is tie the boat. Then he can either grab the fishing rod or the icebox (two things). Then he grabs whatever is left, then he heads home. It seems to me that there are only two ways of performing event C, not six.
Added: Event C has recently been edited to give 2 ways, in which case I agree.
Your method for the second part of the question works (it doesn't have to be in Event A or C; it could be something in Event B as well). Or you can add a thing that can either be done or not, e.g., "wear or not wear a cap."
Best Answer
Roo travels $200$ feet to reach the finish line.
Since Tigger has to travel in $3$ foot jumps, Tigger needs to travel $102$ feet until (s)he can turn around. Then Tigger needs to travel another $102$ feet to reach the start/finish line, a total of $204$ feet. The two jumpers travel at the same rate. So Roo is $4$ feet ahead of Tigger at the end, sort of.
This is in fact not necessarily the case, because it relies on a specific model of jumping. Since we were not given a detailed description of the jump processes of either creature, we are forced to assume something. We implicitly assumed that even although the travelling is divided into jumps, the actual speeds are constant.
Suppose instead that a jump consists of standing in one place flexing the leg muscles for a while, and perhaps having a drink, while the movement part of a jump is quick. Then the answer need not be $4$ feet.