The problem asks us to find the intersection of set $A$ and the union of the complement of set $B$ ($B'$) and set $C$. We are given the universal set $\mu = \{x: 0 \le x \le 6\}$, which implies $\mu = \{0, 1, 2, 3, 4, 5, 6\}$. We are also given $A = \{0, 2, 4, 6\}$, $B = \{1, 2, 3, 4\}$, and $C = \{1, 3\}$.

1. Problem Description

The problem asks us to find the intersection of set $A$ and the union of the complement of set $B$ ($B'$) and set $C$. We are given the universal set $\mu = \{x: 0 \le x \le 6\}$, which implies $\mu = \{0, 1, 2, 3, 4, 5, 6\}$. We are also given $A = \{0, 2, 4, 6\}$, $B = \{1, 2, 3, 4\}$, and $C = \{1, 3\}$.

2. Solution Steps

First, we need to find the complement of set $B$, denoted as $B'$. The complement of $B$ contains all elements in the universal set $\mu$ that are not in $B$.

$B = \{1, 2, 3, 4\}$, so $B' = \mu - B = \{0, 5, 6\}$.

Next, we need to find the union of $B'$ and $C$, which is $B' \cup C$. This means we combine all the elements in $B'$ and $C$ into a single set, removing any duplicates.

$B' = \{0, 5, 6\}$ and $C = \{1, 3\}$, so $B' \cup C = \{0, 1, 3, 5, 6\}$.

Finally, we need to find the intersection of $A$ and $(B' \cup C)$, which is $A \cap (B' \cup C)$. This means we find the elements that are common to both $A$ and $(B' \cup C)$.

$A = \{0, 2, 4, 6\}$ and $B' \cup C = \{0, 1, 3, 5, 6\}$, so $A \cap (B' \cup C) = \{0, 6\}$.

3. Final Answer

$\{0, 6\}$