Five. (Full marks 16 points) Xiao Ming has 5 RMB notes, with a total face value of 20 yuan.
(1) The face values of Xiao Ming's 5 RMB notes are $\qquad$ yuan, $\qquad$ yuan, $\qquad$ yuan, $\qquad$ yuan.
(2) Xiao Ming went to a fruit shop and weighed $x$ pounds of apples ($x$ is an integer), and according to the marked price, he should pay $y$ yuan, which is exactly equal to the sum of the face values of 2 of the 5 RMB notes; at this point, there were still 6 pounds of apples left in the basket. The shop owner then said to Xiao Ming: “If you buy the rest as well, then including what you have already weighed, you can pay 10 yuan in total.” Xiao Ming calculated that this would reduce the price per pound by 0.5 yuan compared to the original marked price, and in the spirit of mutual benefit, he agreed. Try to find $x$ and $y$.
(1) Xiaoming's 5 RMB notes have denominations of 1 yuan, 2 yuan, 2 yuan, 5 yuan, and 10 yuan.

It is easy to see that, among the 5 RMB notes, the smallest denomination cannot be less than "1 yuan." For example, if the smallest denomination is "1 jiao," at least 3 more notes would be needed to make up a whole 1 yuan, and the remaining 1 note cannot be 19 yuan.
The smallest denomination being 2 jiao or 5 jiao can also be easily ruled out, so the smallest denomination cannot be less than "1 yuan."
The smallest denomination being 2 yuan or 5 yuan can also be easily ruled out.
Therefore, the denominations of the 5 RMB notes can only be 1 yuan, 2 yuan, 2 yuan, 5 yuan, and 10 yuan.
(2) From (1), we know that when 2 notes are taken from Xiaoming's 5 RMB notes, the sum being less than 10 yuan has only 4 cases: $1+2=3, 2+2=4, 1+5=6, 2+5=7$, i.e., the value of $y$ can only be $3, 4, 6, 7$.
If $y=3$, then the original price is $\frac{3}{x}$ yuan, and the marked price is $\frac{10}{x+6}$ yuan, giving $\frac{10}{x+6}=\frac{3}{x}-0.5$, which leads to $x^{2}-20 x-36=0$, and it is easy to see that there is no integer solution;
If $y=4$, we get $x^{2}+18 x-48=0$, with no integer solution;
If $y=6$, we get $x^{2}+14 x-72=0$, solving it gives $x_{1}=-18$ (which does not meet the problem's requirements, discard), $x_{2}=4$;
If $y=7$, we get $x^{2}+12 x-84=0$, with no integer solution.
Therefore, the solution to this problem is $x=4, y=6$.