#### Answer

shown below

#### Work Step by Step

(a)
The set\[\left\{ 0,1,2,3,4,5,6 \right\}\]is closed under the operation of clock addition, because the entries of the table are all the elements of the set.
Hence, the set \[\left\{ 0,1,2,3,4,5,6 \right\}\]is closed under the operation of clock addition.
(b)
Consider the expression of LHS.
\[\begin{align}
& \left( 3\oplus 5 \right)\oplus 6=1\oplus 6 \\
& =0
\end{align}\]
Now, find RHS as:
\[\begin{align}
& 3\oplus \left( 5\oplus 6 \right)=3\oplus 4 \\
& =0
\end{align}\]
Hence, \[\left( 3\oplus 5 \right)\oplus 6=3\oplus \left( 5\oplus 6 \right)\]. Associative property has been verified.
(c)
Identity element is that element, which does not change anything when clock addition is used.
Now, as we can see that,
\[0\oplus 0=0,\ 0\oplus 1=1,\ 0\oplus 2=2,\ 0\oplus 3=3,\ 0\oplus 4=4,\ 0\oplus 5=5,\ 0\oplus 6=6\]and,
\[0\oplus 0=0,\ 1\oplus 0=1,\ 2\oplus 0=2,\ 3\oplus 0=3,\ 4\oplus 0=4,\ 5\oplus 0=5,\ 6\oplus 0=6\]
Thus, 0 is the identity element.
Hence, 0 is the identity element in the 7-hour clock addition.
(d)
When an element is added to its inverse, the result is the identity element. The identity element is 0. Find inverse of each element as,
Let \[x\]be the inverse of any element, such that
\[x\oplus 0=0\], and from the table \[0\oplus 0=0\]
Thus, 0 is the inverse of element 0.
\[x\oplus 1=0\], and from the table \[6\oplus 1=0\]
Thus, 6 is the inverse of element 1.
\[x\oplus 2=0\], and from the table \[5\oplus 2=0\]
Thus, 5 is the inverse of element 2.
\[x\oplus 3=0\], and from the table \[4\oplus 3=0\]
Thus, 4 is the inverse of element 3.
\[x\oplus 4=0\], and from the table \[3\oplus 4=0\]
Thus, 3 is the inverse of element 4.
\[x\oplus 5=0\], and from the table \[2\oplus 5=0\]
Thus, 2 is the inverse of element 5.
\[x\oplus 6=0\], and from the table \[1\oplus 6=0\]
Thus, 1 is the inverse of element 6.
Hence, inverse of elements 0,1,2,3,4,5 and 6 are 0,6,5,4,3,2 and 1, respectively.
(e)
\[4\oplus 5=2\]
Now, find RHS as,
\[5\oplus 4=2\]
And,
\[6\oplus 1=0\]
Now, find RHS as,
\[1\oplus 6=0\]
Both the equations follow commutative property.
Hence,\[4\oplus 5=5\oplus 4\]and\[6\oplus 1=1\oplus 6\]follows commutative property and has been verified.