For this solution we have to thank Andrei from Tudor Vianu
National College, Bucharest, Romania.
(a) Evaluating $(x + x^2 + x^3 + x^4)^2$, I obtained $$x^2
+2x^3+3x^4+4x^5+3x^6+2x^7+x^8$$
(b) The sum of two spinners labelled 1, 2, 3 and 4 varies from
2 to 8. There are 16 possibilities in total, giving the following
probabilities: \begin{eqnarray} &\text{Scores} &2 &3
&4 &5 &6 &7 &8 \\ &\text{Frequencies}
&1 &2 &3 &4 &3 &2 &1 \\
&\text{Probabilities} &0.0625 &0.125 &0.1875
&0.25 &0.1875 &0.125 &0.0625 \end{eqnarray}
(c) I observe that the frequency distribution is the same as
the coefficients from the expansion of the polynomial from
(a).
(d) Using the computer simulation, I took two spinners
labelled from 1 to 4, obtaining the following table of
frequencies:
Exp.$\quad$ 
Relative frequency$\quad$ 
0 
0 
1 
0 
2 
0.0631 
3 
0.1253 
4 
0.1872 
5 
0.2493 
6 
0.188 
7 
0.1248 
8 
0.062

The results obtained are very close to the theoretical
frequency distribution of the scores.
(e) $(x + x^2 + x^3 + x^4)^2$ could be factorised as follows:
$x^2 (1+x)^2 (1+x^2)^2$ and these factors can be rewritten
as:
1) $(1+x)(x+x^2) (1+x^2)(x+x^3)$
2) $(x+x^2)^2 (1+x^2)^2$
3) $(1+x)^2(x+x^3)^2$
The first case corresponds to 4 spinners (0,1), (1,2),
(0,2),(1,3), the second to 4 spinners: (1,2), (1,2), (0,2), (0,2)
and the third to another 4 spinners (0,1), (0,1), (1,3), (1,3).\par
(f) Using the simulation, I obtained similar frequency
distributions as with two 1,2,3,4 spinners.
Score$\quad$ 
Frequency$\quad$ 
distributions$\quad$ 


(1) 
(2) 
(3) 
2 
0.3324 
0.3312 
0.3359 
3 
0.4996 
0.5021 
0.4982 
4 
0.1679 
0.1665 
0.1658

The theoretical probabilities would be $2/6 = 33.33 \%$, for
2, $3/6 = 50\%$ for 3, and $1/6 = 16.66\%$, which are very close to
the simulated values.
For a 2spinner, there are equal probabilities of obtaining
the two numbers with which it is labelled. It corresponds to the
polynomial $x^m + x^n$. If $m$ and $n$ are equal, then the
probability ofobtaining that value is 1 ($100\%$).\par For a
3spinner, I have 3 cases:
 All three numbers are different. There is a probability of 1/3
($33.33\%$) of obtaining any of the three numbers. It corresponds
to the polynomial $x^m + x^n + x^p$, with $m \neq n \neq p$
 Two of the three numbers are equal. There is a probability of
1/3 ($33.33\%$) of obtaining the unique number, and a probability
of $66.67\%$ of obtaining the number which appears twice. It
corresponds to the polynomial $x^m + 2x^n$, $m \neq n$.
 All 3 numbers are equal. The probability of obtaining this
number is $100\%$.
Two ordinary dice are equivalent to 2 6spinners. The
theoretical probabilities and the simulated ones are:
Number 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
Theoretical 
1/36 
2/36 
3/36 
4/36 
5/36 
6/36 
5/36 
4/36 
3/36 
2/36 
1/36 
Frequency 
0.0281 
0.0556 
0.083 
0.1123 
0.1397 
0.1674 
0.1381 
0.1093

0.0842

0.0539 
0.028 
This case corresponds to the polynomial $(x + x^2 + x^3 + x^4
+ x^5 + x^6)^2$. The expansion of this polynomial is: $$x^2 + 2x^3
+ 3x^4 + 4x^5 + 5x^6 + 6x^7 + 5x^8 + 4x^9 + 3x^{10} + 2x^{11} +
x^{12}.$$ This polynomial could be factorised as follows: $$(x +
x^2 + x^3 + x^4 + x^5 + x^6)^2 = x^2 (1+x)^2 (1+x^2+x^4)^2$$ and
from this decomposition different combinations could be obtained,
all of them producing the same frequency distribution.