### Part (a)
1. **Assume $G$ has property $P$**: Let $f$ be an automorphism of $G$. By property $P$, there exist automorphisms $g$ and $h$ such that for all $x \in G$, $f(x) = g(x) \cdot h(x)$.

2. **Consider the identity element $e \in G$**: Since $f$ is an automorphism, $f(e) = e$. Therefore, $e = g(e) \cdot h(e)$. Since $g$ and $h$ are automorphisms, $g(e) = e$ and $h(e) = e$.

3. **Apply $f$ to the product of two elements**: Let $x, y \in G$. Then,
$$f(x \cdot y) = g(x \cdot y) \cdot h(x \cdot y).$$
Since $g$ and $h$ are automorphisms, we have
$$g(x \cdot y) = g(x) \cdot g(y) \quad \text{and} \quad h(x \cdot y) = h(x) \cdot h(y).$$
Therefore,
$$f(x \cdot y) = (g(x) \cdot g(y)) \cdot (h(x) \cdot h(y)).$$

4. **Use the fact that $f$ is an automorphism**: Since $f$ is an automorphism, it preserves the group operation, so
$$f(x \cdot y) = f(x) \cdot f(y).$$
Substituting the expressions for $f(x)$ and $f(y)$, we get
$$f(x \cdot y) = (g(x) \cdot h(x)) \cdot (g(y) \cdot h(y)).$$

5. **Compare the two expressions for $f(x \cdot y)$**: We have
$$(g(x) \cdot g(y)) \cdot (h(x) \cdot h(y)) = (g(x) \cdot h(x)) \cdot (g(y) \cdot h(y)).$$
By the associativity of the group operation, this implies
$$g(x) \cdot g(y) \cdot h(x) \cdot h(y) = g(x) \cdot h(x) \cdot g(y) \cdot h(y).$$

6. Simplify the equation: Since $g(x)$ and $h(x)$ are elements of $G$, we can cancel $g(x)$ and $h(x)$ from both sides, yielding
$$g(y) \cdot h(y) = h(y) \cdot g(y).$$
This shows that $g(y)$ and $h(y)$ commute for all $y \in G$.

7. **Conclude that $G$ is commutative**: Since $g$ and $h$ are arbitrary automorphisms and $y$ is an arbitrary element of $G$, it follows that $G$ is commutative.

### Part (b)
1. **Assume $G$ is a commutative finite group of odd order**: Let $|G| = 2k + 1$ for some integer $k$.

2. **Consider the automorphism $f$ defined by $f(x) = x^{-1}$**: Since $G$ is commutative, $f$ is an automorphism.

3. **Assume $G$ has property $P$**: Then there exist automorphisms $g$ and $h$ such that for all $x \in G$,
$$f(x) = g(x) \cdot h(x).$$
Therefore,
$$x^{-1} = g(x) \cdot h(x).$$

4. **Consider the element $x = e$**: For the identity element $e \in G$, we have $e^{-1} = e$. Therefore,
$$e = g(e) \cdot h(e).$$
Since $g$ and $h$ are automorphisms, $g(e) = e$ and $h(e) = e$.

5. **Consider the element $x = a$ where $a \neq e$**: Since $G$ is commutative and of odd order, every element $a \in G$ has an inverse $a^{-1} \neq a$. Therefore,
$$a^{-1} = g(a) \cdot h(a).$$

6. **Use the fact that $G$ is commutative**: Since $G$ is commutative, we have
$$a^{-1} = g(a) \cdot h(a) = h(a) \cdot g(a).$$

7. **Consider the order of $a$**: Since $G$ is of odd order, the order of any element $a \in G$ must be odd. Therefore, $a^{-1} \neq a$.

8. **Conclude that $G$ does not have property $P$**: Since $a^{-1} \neq a$ and $g(a) \cdot h(a) = a^{-1}$, it follows that $G$ does not have property $P$.

### Part (c)
1. **Assume $G$ is a finite group of order $4n + 2$**: Let $|G| = 4n + 2$ for some integer $n$.

2. **Consider the automorphism $f$ defined by $f(x) = x^{-1}$**: Since $G$ is a finite group, $f$ is an automorphism.

3. **Assume $G$ has property $P$**: Then there exist automorphisms $g$ and $h$ such that for all $x \in G$,
$$f(x) = g(x) \cdot h(x).$$
Therefore,
$$x^{-1} = g(x) \cdot h(x).$$

4. **Consider the element $x = e$**: For the identity element $e \in G$, we have $e^{-1} = e$. Therefore,
$$e = g(e) \cdot h(e).$$
Since $g$ and $h$ are automorphisms, $g(e) = e$ and $h(e) = e$.

5. **Consider the element $x = a$ where $a \neq e$**: Since $G$ is of order $4n + 2$, there exists an element $a \in G$ such that $a^{-1} \neq a$. Therefore,
$$a^{-1} = g(a) \cdot h(a).$$

6. **Use the fact that $G$ is of order $4n + 2$**: Since $G$ is of order $4n + 2$, the order of any element $a \in G$ must be even. Therefore, $a^{-1} \neq a$.

7. **Conclude that $G$ does not have property $P$**: Since $a^{-1} \neq a$ and $g(a) \cdot h(a) = a^{-1}$, it follows that $G$ does not have property $P$.