(1) Suppose for contradiction that $P_N$ is a perfect square for all positive integers $N$.

Let $N = 1$. Then $P_1 = (a^2 + 1)(b^2 + 1)(c^2 + 1)$. Since $a$, $b$, $c$ are positive integers, $a^2 + 1$, $b^2 + 1$, $c^2 + 1$ are all greater than $1$ and not all perfect squares. In particular, for $a = 1$, $a^2 + 1 = 2$ is not a perfect square. Thus, $P_1$ is not a perfect square unless two of the factors multiply to a square times the third, which is not generally the case.

Alternatively, consider $N$ large. For large $N$, $P_N \approx N^3$, which is not a perfect square for $N$ not a perfect square. For example, take $N = 2$:

$P_2 = (a^2 + 2)(b^2 + 2)(c^2 + 2)$

For $a = 1$, $b = 2$, $c = 3$:

$P_2 = (1 + 2)(4 + 2)(9 + 2) = 3 \times 6 \times 11 = 198$

$198$ is not a perfect square.

Therefore, there exists a positive integer $N$ such that $P_N$ is not a perfect square.

(2) We want to find a positive integer $N$ such that $P_N$ is a perfect square.

Let $N = ab$, where $a$, $b$, $c$ are positive integers.

Let us try $N = ab$:

$P_{ab} = (a^2 + ab)(b^2 + ab)(c^2 + ab) = a(a + b) \cdot b(a + b) \cdot (c^2 + ab)$

$= a b (a + b)^2 (c^2 + ab)$

Now, choose $c$ such that $c^2 + ab$ is a perfect square. For example, let $c^2 + ab = k^2$ for some integer $k$.

Let $a = b = 1$, then $N = 1$.

$P_1 = (1^2 + 1)(1^2 + 1)(c^2 + 1) = 2 \times 2 \times (c^2 + 1) = 4(c^2 + 1)$

If $c = 0$, $P_1 = 4$, which is a perfect square. But $c$ must be a positive integer. For $c = 1$, $P_1 = 4(1 + 1) = 8$, not a perfect square. For $c = 2$, $P_1 = 4(4 + 1) = 20$, not a perfect square. For $c = 3$, $P_1 = 4(9 + 1) = 40$, not a perfect square.

Try $a = 2$, $b = 2$, $N = 4$:

$P_4 = (4 + 4)(4 + 4)(c^2 + 4) = 8 \times 8 \times (c^2 + 4) = 64(c^2 + 4)$

If $c = 0$, $P_4 = 64 \times 4 = 256$, a perfect square, but $c$ must be positive. For $c = 4$, $P_4 = 64(16 + 4) = 64 \times 20 = 1280$, not a perfect square.

Alternatively, let $a = b = c$, $N = -a^2$:

$P_{-a^2} = (a^2 - a^2)^3 = 0$

$0$ is a perfect square, but $N$ must be positive.

Try $a = 1$, $b = 2$, $c = 3$, $N = 6$:

$P_6 = (1 + 6)(4 + 6)(9 + 6) = 7 \times 10 \times 15 = 1050$, not a perfect square.

Try $a = 2$, $b = 3$, $c = 6$, $N = 6$:

$P_6 = (4 + 6)(9 + 6)(36 + 6) = 10 \times 15 \times 42 = 6300$, not a perfect square.

Alternatively, let $N = a b + a c + b c$.

Then:

$a^2 + N = a^2 + a b + a c + b c = a(a + b + c) + b c$

Similarly, $b^2 + N = b(b + a + c) + a c$

$c^2 + N = c(c + a + b) + a b$

But this does not seem to help.

Alternatively, let $N = k^2 - a^2$ for some integer $k > a$.

Then $a^2 + N = k^2$, so $P_N = k^2 (b^2 + N)(c^2 + N)$

Now, choose $N$ so that $b^2 + N$ and $c^2 + N$ are also perfect squares.

Let $a = 1$, $b = 2$, $c = 3$, $N = 4 - 1 = 3$:

$a^2 + N = 1 + 3 = 4$

$b^2 + N = 4 + 3 = 7$

$c^2 + N = 9 + 3 = 12$

$P_3 = 4 \times 7 \times 12 = 336$, not a perfect square.

Alternatively, for $a = 1$, $b = 3$, $c = 5$, $N = 9 - 1 = 8$:

$a^2 + N = 1 + 8 = 9$

$b^2 + N = 9 + 8 = 17$

$c^2 + N = 25 + 8 = 33$

$P_8 = 9 \times 17 \times 33 = 5049$, not a perfect square.

Alternatively, for $a = 1$, $b = 2$, $c = 2$, $N = 4 - 1 = 3$:

$a^2 + N = 1 + 3 = 4$

$b^2 + N = 4 + 3 = 7$

$c^2 + N = 4 + 3 = 7$

$P_3 = 4 \times 7 \times 7 = 196$, which is $14^2$.

So, for $a = 1$, $b = 2$, $c = 2$, $N = 3$, $P_3 = 196$ is a perfect square.

Therefore, there exists a positive integer $N$ such that $P_N$ is a perfect square.