Draw a table for multiplication in modulo 8 on the set 1, 3, 5, 7. Using the table: (i) evaluate (5 3) 7; (ii) find the truth set of m (m + 2) = 3.

The operation is $a \otimes b = (a \times b) \pmod{8}$.

$$\begin{array}{|c|c|c|c|c|} \hline \otimes & 1 & 3 & 5 & 7 \\ \hline 1 & 1 & 3 & 5 & 7 \\ \hline 3 & 3 & 1 & 7 & 5 \\ \hline 5 & 5 & 7 & 1 & 3 \\ \hline 7 & 7 & 5 & 3 & 1 \\ \hline \end{array}$$

First, find $5 \otimes 3$ from the table.

$$5 \otimes 3 = 7$$

Next, evaluate $7 \otimes 7$.

$$7 \otimes 7 = 1$$

So, $(5 \otimes 3) \otimes 7 = \boxed{1}$.

We test each value of $m$ from the set $\{1, 3, 5, 7\}$. Remember that $(m+2)$ is also evaluated modulo 8.

• For $m=1$: $m+2 = 1+2 = 3$. $1 \otimes 3 = 3$. This matches the equation.

• For $m=3$: $m+2 = 3+2 = 5$. $3 \otimes 5 = 7$. This does not match 3.

• For $m=5$: $m+2 = 5+2 = 7$. $5 \otimes 7 = 3$. This matches the equation.

• For $m=7$: $m+2 = 7+2 = 9$. $9 \pmod{8} = 1$. So we evaluate $7 \otimes 1$. $7 \otimes 1 = 7$. This does not match 3.

The values of $m$ that satisfy the equation are 1 and 5.
The truth set is $\boxed{\{1, 5\}}$.