Well done to Navjot from Sherborne Qatar, who sent in a full solution:
I will firstly divide the given bar charts into 2 groups. Graphs A, B, C, and D are made by the sum of the values on the spinners and graphs E, F, G, and H are made by the difference of the values of the spinners.

We know this because of the fact that the x-values of the first 4 graphs start with 2 while the last 4 graphs start with a 0.

Now, I shall determine the spinners used to make these frequency graphs.

Each graph seems to be a sum or a difference of two even or two odd numbers but not an even and an odd number since the final x-value is an even number.

I assume that the spinners are fair (as we will see, the probability of the sums shown on the graph are extremely close to the calculated probability, which is bound to happen after 5000 spins, if there would still be an imbalance in the probability of getting certain values then the spinners would definitely be biased).

As shown in the images, in the case of graph A, the only possible pairs we could use to get a maximum sum of 8 are 1 & 7, 2 & 6, 3 & 5, or 4 & 4. Through the table, I have shown what is the chance getting a certain sum. The numbers shown on each spinner are shown in blue at the left and bottom of the table, and the sums are shown in black inside the table. 
There are 16 black numbers in the table, so 16 possible pairs of numbers on the spinners. Of the sums, 2 only appears once (because to get a sum of 2, you would have to get 1, 1). So the probability of getting 2 is $\frac1{16}$. However, 3 appears twice (1+2 and 2+1) and 4 appears 3 times (1+3 and 2+2 and 3+1), so the probabilities of getting 3 and 4 are $\frac2{16}$ and $\frac3{16}$ respectively.
Through trying the combinations, we would notice that as the difference between the maximum values of the two spinners decrease, the probabilities of getting certain sums start to decrease at a different rate to others (which would create that peak in the graph rather than the plateau). If we were to graph the results, the graph A would most resemble the results of the spinners with the maximum values 4 and 4.

 
Using that logic (that the smaller the difference between the values, the more the graph would represent the normal distribution curve) [that is, that when the spinners both have the same numbers, the graph goes smoothly up and then smoothly down, peaking in the middle] I deduced that graph B was made by using the spinners with the highest values 7 and 7.

However, in graph C and D, we notice a plateau. This suggests that the top values of the two spinners were not equal. For graph C, I shall assume that the probabilities to get any sum between 4 and 10 are equal (since the variation is negligible), and for D, I shall assume the probabilities to get anything between 7 and 11 are equal.

Applying the same method as I used to find the values used in graph A and B, I find out that the spinners used in graph C went up to the values 3 and 9.

	1	2	3	4	5	6	7	8	9
1	2	3	4	5	6	7	8	9	10
2	3	4	5	6	7	8	9	10	11
3	4	5	6	7	8	9	10	11	12

The probabilities ($\text P$) of getting the sums were as follows:
$\text P(2) =\text P(12) = \frac1{27}$
$\text P(3) =\text P(11) = \frac2{27}$
$\text P(4) =\text P(5) =\text P(6) =\text P(7) =\text P(8) =\text P(9) =\text P(10) = \frac3{27} = \frac1{9}$

To produce graph D, spinners of values 6 and 10 would be used.

	1	2	3	4	5	6	7	8	9	10
1	2	3	4	5	6	7	8	9	10	11
2	3	4	5	6	7	8	9	10	11	12
3	4	5	6	7	8	9	10	11	12	13
4	5	6	7	8	9	10	11	12	13	14
5	6	7	8	9	10	11	12	13	14	15
6	7	8	9	10	11	12	13	14	15	16

The probabilities ($\text P$) of getting the sums were as follows:
$\text P(2) =\text P(16) =\frac1{60}$
$\text P(3) =\text P(15) = \frac2{60} = \frac1{30}$
$\text P(4) =\text P(14) = \frac3{60} = \frac1{20}$
$\text P(5) =\text P(13) = \frac4{60} = \frac1{15}$
$\text P(6) =\text P(12) = \frac5{60} = \frac1{12}$
$\text P(7) =\text P(8) =\text P(9) =\text P(10) =\text P(11) = \frac6{60} = \frac1{10}$

The final four graphs would require the same procedure of producing a table with the possible scores on the sides of the tables, but instead of plugging in the sums, we plug in the differences.

Graph E would be made by using spinners with the maximum values of 5 & 5

	1	2	3	4	5
1	0	1	2	3	4
2	1	0	1	2	3
3	2	1	0	1	2
4	3	2	1	0	1
5	4	3	2	1	0

The probabilities ($\text P$) of getting the differences were as follows:
$\text P(0) = \frac5{25} = \frac15$
$\text P(1) = \frac8{25}$
$\text P(2) = \frac3{25}$
$\text P(3) = \frac4{25}$
$\text P(4) = \frac2{25}$
Since $\text P(5) = 0$ the results here match the graph E.

Graph F would be made by using spinners with the maximum values of 9 & 9

	1	2	3	4	5	6	7	8	9
1	0	1	2	3	4	5	6	7	8
2	1	0	1	2	3	4	5	6	7
3	2	1	0	1	2	3	4	5	6
4	3	2	1	0	1	2	3	4	5
5	4	3	2	1	0	1	2	3	4
6	5	4	3	2	1	0	1	2	3
7	6	5	4	3	2	1	0	1	2
8	7	6	5	4	3	2	1	0	1
9	8	7	6	5	4	3	2	1	0

The probabilities ($\text P$) of getting the differences were as follows:
$\text P(0) = \frac9{81} = \frac19$
$\text P(1) = \frac{16}{81}$
$\text P(2) = \frac{14}{81}$
$\text P(3) = \frac{12}{81} = \frac4{27}$
$\text P(4) = \frac{10}{81}$
$\text P(5) = \frac8{81}$
$\text P(6) = \frac6{81} = \frac2{27}$
$\text P(7) = \frac4{81}$
$\text P(8) = \frac2{81}$
Since $\text P(9) = 0$ the results here match the graph F.

Graph G would be made by using spinners with the maximum values of 3 & 7

	1	2	3	4	5	6	7
1	0	1	2	3	4	5	6
2	1	0	1	2	3	4	5
3	2	1	0	1	2	3	4

The probabilities ($\text P$) of getting the differences were as follows:
$\text P(0) = \frac3{21} = \frac17$
$\text P(1) = \frac5{21}$
$\text P(2) = \frac4{21}$
$\text P(3) =\text P(4) = \frac3{21}$
$\text P(5) = \frac2{21}$
$\text P(6) = \frac1{21}$
Since $\text P(7) = 0$ the results here match the graph G.

Graph H would be made by using spinners with the maximum values of 10 & 4

	1	2	3	4	5	6	7	8	9	10
1	0	1	2	3	4	5	6	7	8	9
2	1	0	1	2	3	4	5	6	7	8
3	2	1	0	1	2	3	4	5	6	7
4	3	2	1	0	1	2	3	4	5	6

The probabilities ($\text P$) of getting the differences were as follows:
$\text P(0) = \frac4{40} = \frac1{10}$
$\text P(1) = \frac7{40}$
$\text P(2) = \frac6{40} = \frac3{20}$
$\text P(3) = \frac5{40} = \frac18$
$\text P(4) =\text P(5) =\text P(6) = \frac4{40} = \frac1{10}$
$\text P(7) = \frac3{40}$
$\text P(8) = \frac2{40} = \frac1{20}$
$\text P(9) = \frac1{40}$
Since $\text P(10) = 0$ the results here match the graph F.

In conclusion, these are the spinners used for each graph:
A: 4 & 4
B: 7 & 7
C: 3 & 9
D: 6 & 10
E: 5 & 5
F: 9 & 9
G: 3 & 7
H: 10 & 4

Final challenge
This is Navjot's solution to the final challenge. Navjot's descriptions of the shapes of the graphs are good, but one of the numbers is not quite right.
The sum of two 1-30 spinners: The bar charts would be similar to graph A or B as it peaks at 16, ((the maximum x-value)$\div$2)$+$1.

The difference between two 1-20 spinners:  The bar charts would peak in a similar way to graph E and F. It would peak at 1 and then keep decreasing up to 9, ((the maximum x-value)$\div$2)$-$1. In fact, they would decrease up to 19, because the largest difference you could get would be between 20 and 1: a difference of 19.

The sum of a 1-20 and a 1-30 spinner: The bar chart would be similar to graphs C and D. The probability would rise up to 21 and then stay constant till 31 and then decrease (the plateau between 21 and 31 was deduced by the fact that the plateau seems to be between the sum that are 1 more than the maximum value of either spinner (when maximum values of spinners are 3 and 9, the graph plateaus between 4 and 10).

The difference between a 1-20 and a 1-30 spinner: Similar to graphs G and H, this bar chart would peak at 1 and then keep decreasing up to 29 which would be the largest difference possible difference.