### Part (a)

1. Graph Construction: Construct an undirected graph $G$ where each vertex corresponds to an element in $T$. There is an edge between vertices $s$ and $t$ if and only if $\gcd(s, t) > 1$.

2. **Non-Good Subset $X_0$**: Since $X_0$ is not a good subset, there exists at least one vertex $v \in T$ such that $v$ is not adjacent to any vertex in $X_0$. This means $\gcd(v, x) = 1$ for all $x \in X_0$.

3. Counting Pairs: Consider the set of pairs $(X_0, Y)$ in $A$. For each subset $Y \subseteq T$, we need $\gcd(x, y) = 1$ for all $x \in X_0$ and $y \in Y$.

4. **Splitting by $v$**: Since $v$ is not adjacent to any vertex in $X_0$, $v$ can either be included in $Y$ or not. This creates a natural partition of the pairs $(X_0, Y)$ into two disjoint sets: those that include $v$ and those that do not.

5. Even Count: Each pair $(X_0, Y)$ can be uniquely paired with $(X_0, Y \cup \{v\})$ if $v \notin Y$. This implies that the total number of such pairs is even.

Thus, if $X_0$ is not a good subset, the number of pairs $(X_0, Y)$ in $A$ is even.

### Part (b)

1. **Set $A$**: Consider the set $A = \{(X, Y) \mid X \subseteq T, Y \subseteq T, \gcd(x, y) = 1 \text{ for all } x \in X, y \in Y\}$.

2. Pairing Property: For every pair $(X, Y) \in A$, the pair $(Y, X) \in A$ as well. If $X = Y$, then $\gcd(x, y) = 1$ for all $x, y \in X$, which implies $X$ and $Y$ must be empty (since $\gcd(x, x) = x > 1$ for any $x \in T$).

3. **Odd Cardinality of $A$**: Since each pair $(X, Y)$ can be paired with $(Y, X)$ and the only unpaired element is $(\emptyset, \emptyset)$, the total number of elements in $A$ is odd.

4. Counting Good Subsets: For each subset $X \subseteq T$, count the number of times it appears as the first element of a pair in $A$:
- If $X$ is a good subset, it appears an odd number of times.
- If $X$ is not a good subset, it appears an even number of times (as shown in part (a)).

5. Conclusion: Since the total number of pairs in $A$ is odd, the number of good subsets must be odd to balance the count.

$\blacksquare$