Bginner

Try to write $1/a^2+1/b^2$ as an expression involving only $a+b$ and $ab$.  The values of $a+b$ and $ab$ can be obtained from the coefficients of $5x^2-11x+4$ almost by inspection. (Vieta)

Note that $(m + 1/m)^3 = m^3 + 3m + 3/m + 1/m^3 = (m^3 + 1/m^3) + 3(m+1/m)$.  So $m^3+1/m^3 = (m+1/m)^3 - 3(m+1/m) =8^3-3*8$

First, count how many paths goes from C to A without going through B.  Call this number n1.
Next, count how many paths goes from D to A without going through B.  Call this number n2.
The answer is n1 + n2.
 

Let $k = \sqrt2+\sqrt5$.  We first compute $k^2$, $k^3$, $k^4$.  They are
\begin{eqnarray*}
    k^2 &=& 2\sqrt{10}+7 \\
    k^3 &=& 11\sqrt5+17\sqrt2 \\
    k^4 &=& 28\sqrt{10}+89.
\end{eqnarray*}
Let the polynomial be $P(x)=x^4+bx^3+cx^2+dx+e$.   We have
\begin{eqnarray*}
    P(k) &=& k^4 + bk^3+ck^2+dk+e \\
    &=& (28\sqrt{10}+89) + (11\sqrt5+17\sqrt2)b + (2\sqrt{10}+7)c + (\sqrt2+\sqrt5)d + e \\
    &=& (28 + 2c) \sqrt{10} + (11b + d)\sqrt5 + (17b + d)\sqrt2 + (89 + 7c + e).
\end{eqnarray*}
The only way for $P(k)$ to be 0 is that
\begin{eqnarray*}
    28 + 2c &=& 0 \\
    11b + d &=& 0 \\
    17b + d &=& 0 \\
    89 + 7c + e &=& 0
\end{eqnarray*}
Solving the above system of equations gives $b=0$, $c=-14$, $d=0$, and $e=9$.
Hence, the disired polynomial is $P(x) = x^4 - 14x^2 + 9$.

Divide the polynomial $x^3 -2x^2$ by the polynomial $x^2-x-1$.   The remainder is the answer.   Do you see why this is correct?
 

Let the polynomial $P$ be defined by $P(x)=x^3+bx^2+cx+d$.   Now substitute $\sqrt[3]{4}+1$ for $x$ in the formula for $P(x)$ to get $P(\sqrt[3]{4}+1)$ and collect terms so that you have
\[
    P(\sqrt[3]{4}+1) = k\cdot 4^{1/3} + \ell\cdot 4^{2/3} + m
\]
where $k, \ell, m$ are integral expressions involving $b,c,d$.  Now set up three equations $k=0$, $\ell=0$, and $m=0$, and solve for $b,c,d$.   You should get the solution $P(x) = x^3-3x^2+3x-5$.   (You should check that $P(\sqrt[3]{4}+1) = 0$.)

The number a in such context is always taken to be a nonnegative integer less than m.  If you find a solution that is negative, do add an appropriate multiple of m to it to make it nonnegative and less than m.  E.g. if you find a solution of -2 for a, and your m is 5, then add 5 to a to get -2+5 = 3.  So your a is actually 3.   So a+m in this case is 3+5 = 8.

Isn't N a four-digit positive integer?

Well, will it help to know that the desired smallest positive x is in the interval (1.1, 1.2)?

We'll set up an equation by "counting the same thing in two different ways."  Consider the process of tracing out the star by following the five segments in this order a-c, c-e, e-b, b-d, d-a and adding up all the numbers we encountered in the process.

One way of adding the numbers goes like this: Each time we draw a segment we encounter four numbered points.   The picture clearly shows that each numbered point is the intersection of two segments.  So each number is added to the total exactly twice.  So the total sum we get is 2(a+b+c+d+e) + 2(2+3+4+5+6).

Another way of adding the numbers goes like this: We draw the stars by drawing five segments and each segment encountered contains four numbers adding up to 28.   So the total sum we get is 5*28.

The sums we get in two different ways must be the same, so now we can set up an equation

    2*(a+b+c+d+e) + 2*(2+3+4+5+6) = 5*28.

Solving, we get

    a+b+c+d+e = (5*28 - 2*(2+3+4+5+6)) / 2.