Solution (1)
$$\begin{aligned} & (x+y+z)^{3}-27 x y z \\ = & x^{3}+y^{3}+z^{3}+3\left(x^{2} y+y^{2} z+z^{2} x\right)+3\left(x y^{2}+y z^{2}+z x^{2}\right)+6 x y z \\ & -27 x y z \\ = & (x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)+3\left(x^{2} y+y^{2} z+\right. \\ & \left.z^{2} x+x y^{2}+y z^{2}+z x^{2}-6 x y z\right) \\ & =\frac{x+y+z}{2}\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]+3\left[x(y-z)^{2}+\right. \\ & \left.y(z-x)^{2}+z(x-y)^{2}\right] . \end{aligned}$$

Assume without loss of generality that $x \geqslant y \geqslant z$, then $z \geqslant 1, y \geqslant 1, x \geqslant 2$.
Thus,
$$\begin{aligned} & (x+y+z)^{3}-27 x y z \\ \geqslant & \frac{1+1+2}{2} \cdot\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]+3\left[(y-z)^{2}+\right. \\ & \left.(z-x)^{2}+(x-y)^{2}\right] \\ = & 5 \cdot\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] \\ \geqslant & 10 \end{aligned}$$

Therefore, $(x+y+z)^{3}-27 x y z \geqslant 10$, and equality holds when $(x, y, z)=(2,1,1)$.
(2) Assume without loss of generality that $x>y>z$, then $z \geqslant 1, y \geqslant 2, x \geqslant 3$.
From part (1),
$$\begin{aligned} (x+y+z)^{3}-27 x y z \geqslant & \frac{6}{2} \cdot\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]+3 \cdot[3(y \\ & \left.-z)^{2}+2(z-x)^{2}+(x-y)^{2}\right] \\ \geqslant & 3 \cdot\left(1^{2}+1^{2}+2^{2}\right)+3 \cdot\left(3 \cdot 1^{2}+2 \cdot 2^{2}+1^{2}\right) \\ & =54 \end{aligned}$$

Therefore, $(x+y+z)^{3}-27 x y z \geqslant 54$, and equality holds when $(x, y, z)=(3,2,1)$. Note that part (1) is equivalent to the following proposition:
For distinct positive integers $a, b, c$, we have
$$\frac{a+b+c}{3} \geqslant \sqrt[3]{a b c+\frac{10}{27}}$$

Of course, (1) can also be proven using the following method (incremental substitution):
Proof Assume without loss of generality that $1 \leqslant a \leqslant b \leqslant c, b=a+x, c=a+y$. Then $x, y \geqslant 0$, and $x, y$ are not both $0 (x \leqslant y)$.

At this point, (1) is equivalent to proving:
$$9 a\left(x^{2}-x y+y^{2}\right)+(x+y)^{3} \geqslant 10$$

Since $a \geqslant 1, x^{2}-x y+y^{2} \geqslant 1, x+y \geqslant 1$, the above inequality clearly holds, and equality is achieved when $a=1, x^{2}-x y+y^{2}=1, x+y=1$, i.e., $(a, x, y)=(1,0,1)$.