First of all, do not feel discouraged by you having a hard time to solve these problems. While there are short solutions, and there will be quite some people finding those fairly quickly, it happens very often that a problem is trivial - if you have seen the required method before - or really very hard. In my case, I had to go back to my mental box of methods from the time when I myself took part in such competitions; and even so they took me a bit longer than these 10 minutes.
Also note that, from what I can see here, you are not required to prove your assumptions; which also allows you to save a lot of time. You merely need to make an educated guess. (May I ask what competition they are from? I actually found them quite enjoyable.)
What I am going to do is first give you some pointers to each problem, for you to maybe give it another go and try to figure them out yourself, and then include a solution, where I will also try to illuminate my thought process - often you (or, at least, I) see a proof where at first you have no idea how people thought of that in a reasonable amount of time, so maybe that will help you out as well.
Hints
Try to find some way to factorize your inequality. This is easier if it is homogenous, you can use that $(x+y)=1$. I know this is a bit of a cruel hint as it takes quite some time and practice, but I haven't found anything more pretty than that, sorry.
Your approach works very well, in fact I believe you have done the most part. Now, have you heard of the term 'telescope sum'? If so, try to see what happens if you substitute $n$ with $n+1$.
Can you say something about the roots of the polynomial? If so, take a very close look at what is positive and what is negative, and what exactly is under the root. Compare that to the roots of your possible factors.
Solutions/Thoughts on the problems
Main parts put into spoilers (provided I figured out the way these work correctly), so they don't literally spoil your fun.
1.
In my opinion, the nastiest one.
What I saw when I first looked at the task was along the lines of "Yay, AM-GM", together with the reasonable expectation that the maximum value would be reached for $x=y=\frac{1}{2}$. I tried to prove that, and, in hindsight unsurprisingly, failed. The reason for that lies somewhat in the structure of your term, which I didn't realize at first - if you write down $x^4y+xy^4=xy(x^3+y^3)$ you will notice that these two factors assume their maximum value at very different points; the first when both $x$ and $y$ are equal, the other when $x=1,y=0$. This makes use of means fairly difficult - there might still be a beautiful solution involving these, though, if someone finds one I'd be very interested to see it.
Since I know of no other methods to transform the expression to a more approachable form, the next logical step was to try and factorize. I had actually tried this once before, albeit for a very short time, and dismissed it as probably too complicated.
The kind of factorization you should really get used to seeing in these kinds of tasks is something along the lines of
$x^4y+xy^4=xy(x^3+y^3)=xy(x+y)(x^2-xy+y^2)=^{(x+y=1)}xy(x^2-xy+y^2)$.
What we want is to find an upper bound for this expression, written in an, if possible, homogenous inequality with an expression we can easily calculate. A possible example that fulfills this would be the claim
$\frac{1}{c}=\frac{(x+y)^4}{c}\geq xy(x^2-xy+y^2)$
for some constant $c>0$. In fact, considering the task, we can assume that $c\in \left\{6,8,12,16\right\}$.
Now going along trying to prove that, we can just put everything on one side:
$LHS(c):=x^4+(4-c)x^3y+(6+c)x^2y^2+(4-c)xy^3+y^4\geq 0$,
gained by simple transformations of the above inequality. This would certainly be true if our left hand side was a square. Now, if we look at the structure of this, we can see that we have $x^4,y^4,x^2y^2$, the corresponding inner products, and are symmetrical. Because of that, we could be motivated to look at the following:
$SQR(a):=(x^2+y^2+axy)^2=(x^4+y^4+(a^2+2)xy+2ax^3y+2axy^3)$
for some real $a$. This is what we would like to have, if we can find an $a$ and a $c$ so that $LHS(c)=SQR(a)$, we would have proven that $\frac{1}{c}\geq x^4y+xy^4$ - because, then, $LHS(c)$ would be a perfect square and thus greater than $0$.
But now that we have those two expressions, seeing as they are polynomials, we can simply compare coefficients of $LHS(c)$ and $RHS(a)$. This comparison leads us to $a^2+2=6+c, 2a=4-c$, a system of equations which yields a reasonable solution only for $c=-4$:
We can add both equations, leading to $6+c+4-c=10=a^2+2+2a \Leftrightarrow a^2+2a-8=0 \Rightarrow a\in \left\{2,-4\right\}$.
$a=2$ leads to $4=4-c \Leftrightarrow c=0$, which doesn't really give a great upper bound since $\frac{1}{c}$ wouldn't make a lot of sence then.
$a=-4$, however, leads to $c=12$, which does leave us $\frac{1}{12}$ as our upper bound. Just to make sure we actually assume this maximum, we should check whether our square can actually be $0$, but $(x^2-4xy+y^2)$ takes a positive value for $x=0$ and a negative one for $x=y=\frac{1}{2}$, so that is guaranteed.
Note that instead of forming equations for $c$ one could also have just tried all four options, there might be a pretty argument for why $12$ is the only reasonable try but in that case it has escaped my notice.
So, what we actually did there: We simply put in some upper bound $\frac{1}{c}$. Then we transformed into an expression $LHS(c)\geq 0$, and tried to get this expression to be a perfect square. Thus, we looked at the type of perfect square this could reasonably be, of the form $SQR(a)$, and then compared coefficients trying to find out which values for $a$ are possible and which values for $c$ they correspond to. Once we did that, we gained an equation system giving us a solution for $c=12,a=-4$, giving us $LHS(12)=SQR(-4)\geq 0$, and we are done.
As a final thought, differentiating this problem by hand seems not to be a sensible solution - you are quite fast, true, but I do not think one would actually find the roots of the resulting polynomial. In my case, I would have probably seen that $\frac{1}{2}$ is one and just assumed it to be a maximum - and failed the question miserably.
To summarize:
$\frac{1}{12}=\frac{(x+y)^4}{12}\geq x^4y+xy^4 \Leftrightarrow (x^2+y^2-4xy)^2\geq 0$,
gained by trial. And some thoughts, but mostly trial.
2.
Not much to say here. First thoughts on my mind: expansion into partial fractions. Ah, wait, he already did that.
Now the crucial thought you seemingly missed was "oh, how nice would it be if all those summands could magically disappear. Could I possibly get that somehow?", and "what happens if I put in $n+1$ instead of $n$?"
Let's just take a look at that:
$\frac{1}{(n+1)^2-(n+1)+1}=\frac{1}{n^2+n+1}$.
Ah. That's what happens, and it is wonderful. Because if we just put that into our little sum, we have:
$\frac{1}{2}\sum_{n=1}^{99}{\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}}=\frac{1}{2}[\sum_{n=1}^{99}{\frac{1}{n^2-n+1}}-\sum_{n=2}^{100}{\frac{1}{n^2-n+1}}]=\frac{1}{2}[\frac{1}{1^2-1+1}-\frac{1}{100^2-100+1}]=\frac{1}{2}-\frac{1}{18802}$,
which is quite clearly more than $0.49$.
3.
Since polynomial division is quite an arduous task, we don't want that. It might work, if you're good with calculations it might even work out in a reasonable time, I was too lazy to do it myself and check how much time it actually takes. However, what we can see is that
$P(x):=x^6+10x^3-27=(x^3)^2+10x^3-27 $
is a quadratic polynomial in $x^3$, so we can get its roots:
$\lambda_{1,2}=-5\pm 2\sqrt{13} \quad \Rightarrow\quad x^3\in\left\{\lambda_1,\lambda_2\right\}$.
Note that this also means that we have only $2$ real roots, which are the unique real third roots of our $\lambda_{1,2}$.
Now the interesting part of that is that the roots of a factor polynomials $Q(x)$ must also be roots of $P$ (because $Q(x)=0 \Rightarrow P(x)=Q(x)r(x)=0$), and thus be one of the third roots of one of these $\lambda$s.
Depending on their signs, we get as roots $\pm \frac{1}{2} \pm \frac{i}{2}\sqrt{11}$ (in case the sign before the $3$ is positive) or $\pm\frac{1}{2}\pm\frac{1}{2}\sqrt{13}$ (else).
Now, since we know that the third power of these roots, for any factor polynomial, must be a $\lambda$, we can simply look at what happens if we take these to the third power. We have
$(a+b\sqrt{x})^3=a^3+3ab^2x+(3a^2b+b^3x)\sqrt{x}$ for rational $a,b,x$.
Now imagine our factor polynomial's non-rational part is of the form $\sqrt{11}$. Raising it to the third power, if you look above, the non-rational part is only multiplied by rational numbers, so there is no way to transform $\sqrt{11}$ into $\sqrt{13}$.
Thus, we can rule out the first two possibilities, meaning the non-rational part is $\sqrt{13}$ and the sign of $3$ is $+$. Now we only need to know the sign of $x$ in our polynomial. For this we can look at another aspect of our roots:
We have two $\lambda$s, one negative, and one positve; the absolute value of the positive $\lambda_2$ being less than that of the negative $\lambda_2$. Since $x\mapsto x^3$ is monotonous, the same must be true for the roots of our factor polynomial: Imagine our factor polynomial $Q$ would have a negative root $\psi_1$ with a smaller absolute than its positive $\psi_2$. Then $|\lambda_1|=|\psi_1^3|<|\psi_2^3|=|\lambda_2|$, which does not work out.
So we know that our factor polynomial's positive root must be less than it's negative's absolute, and checking for that we can easily see that the sign of $x$ in the factor polynomial must be $+$, and hence we will go for option c.
Again, don't feel bad if you didn't get these at first. Some might consider these methods standard, and at some point they will be for you as well, but until then such problems are hard. There is no talking around that. So, props for you for trying, and I hope this helped at least somewhat. Keep it up!
Best Answer
$$P_1(x)=\frac{x^{24}-1}{x^2-1}$$ $$P_2(x)=\frac{x^{12}-1}{x-1}$$ $$\frac{P_1(x)}{P_2(x)}=\frac{x^{24}-1}{x^{12}-1}\cdot\frac{x-1}{x^2-1}=\frac{x^{12}+1}{x+1}$$ Then Ruffini's rule tells us that the remainder of this reduced division is the polynomial $x^{12}+1$ evaluated at $-1$, i.e. 2. When the top and bottom of $\frac2{x+1}$ are multiplied by $\frac{x^{12}-1}{x^2-1}$, the denominator becomes $P_2(x)$ and the numerator gives the final answer of $\frac{2(x^{12}-1)}{x^2-1}=2+2x^2+2x^4+2x^6+2x^8+2x^{10}$.