(1) Find the minimum value of the smallest side length of a peculiar triangle;
(2) Prove that there are infinitely many isosceles peculiar triangles;
(3) How many non-isosceles peculiar triangles are there?
(1) Let $a, b, c (a \leqslant b \leqslant c)$ be the side lengths of a peculiar triangle. Then, by Heron's formula, we have
$$
\begin{aligned}
16 \Delta^{2}= & (a+b+c)(a+b-c) . \\
& (a-b+c)(-a+b+c) .
\end{aligned}
$$

Since $(a, b, c)=1$, at least one of $a, b, c$ must be odd. If there is an odd number of odd numbers among $a, b, c$, then $a+b+c, a+b-c, a-b+c, -a+b+c$ are all odd, which contradicts equation (1).
Therefore, exactly two of $a, b, c$ are odd.
If $a=1$, by $c1$, it leads to a contradiction.
When $c=b+1$, $b, c$ are one odd and one even.
Thus, $a, b, c$ have exactly one odd number, which is a contradiction.
If $a=4$, then $b, c$ are both odd.
By $cb-\Delta_{1}$, we have $b+\Delta_{1}=4, b-\Delta_{1}=1$, hence $2b=5$, which is a contradiction.
When $c=b+2$,
$$
p=\frac{1}{2}(a+b+c)=b+3, \\
\Delta^{2}=(b+3)(b-1) \times 3 \times 1,
$$
$$
3 \Delta_{1}^{2}=(b+3)(b-1).
$$

If $3 \mid (b+3)$, then $3 \mid b$, which contradicts the peculiar triangle. If $3 \mid (b-1)$, then $3 \mid (b+2)=c$, which also contradicts the peculiar triangle.
In summary, $a \geqslant 5$.
Moreover, $(5,5,8)$ is a peculiar triangle, so the minimum value of the smallest side length of a peculiar triangle is 5.
(2) If $m, n \in \mathbf{N}_{+}, m>n, 3 \mid m \dot{n}, (m, n)=1, m, n$ are one odd and one even, then $\left(m^{2}+n^{2}, m^{2}+n^{2}, 2 m^{2}-2 n^{2}\right)$ is a peculiar triangle.
In fact, $\Delta=2 m n\left(m^{2}-n^{2}\right)$ is an integer.
Furthermore, since $m, n$ are one odd and one even, then $\left(m^{2}+n^{2}, 2\right)=1$.
Thus, $\left(m^{2}+n^{2}, 2 m^{2}-2 n^{2}\right)$
$=\left(m^{2}+n^{2}, m^{2}-n^{2}\right)=\left(m^{2}+n^{2}, m^{2}\right)$
$=\left(m^{2}, n^{2}\right)=1$.
Finally, because $3 \mid m n$, and $(m, n)=1$, one of $m, n$ is a multiple of 3, so $m^{2}+n^{2}, 2 m^{2}-2 n^{2}$ are not multiples of 3.
In particular, taking $m=6 k+1, n=6$, then
$\left(36 k^{2}+12 k+37,36 k^{2}+12 k+37,72 k^{2}+24 k-70\right)$

is a peculiar triangle.
Similarly, if $m, n \in \mathbf{N}_{+}, m>n, 3 \mid \left(m^{2}-n^{2}\right)$, $(m, n)=1, m, n$ are one odd and one even, then $\left(m^{2}+n^{2}, m^{2}+n^{2}, 4 m n\right)$ is a peculiar triangle.
In particular, taking $m=6 k+1, n=2$, then
$$
\left(36 k^{2}+12 k+5,36 k^{2}+12 k+5,48 k+8\right)
$$

is a peculiar triangle.
(3) There are infinitely many non-isosceles peculiar triangles.
Take $t \equiv 5(\bmod 30)$, and let
$$
\begin{array}{l}
a=5 t^{2}, b=\frac{1}{4}\left(25 t^{4}-6 t^{2}+1\right), \\
c=\frac{1}{4}\left(25 t^{4}+6 t^{2}+1\right) .
\end{array}
$$

Since $t$ is odd, $a, b, c$ are integers, and clearly $a<b<c$.

Also, since $t$ is not a multiple of 3, $a, b, c$ are not multiples of 3.

Finally, since $5 \mid t$, $b, c$ are not multiples of 5, and thus, by $\left(t^{2}, 25 t^{4} \pm 6 t^{2}+1\right)=1$, we have $(a, b, c)=1$.
By calculation, $\Delta=\frac{1}{2} t^{2}\left(25 t^{4}-1\right)$ is an integer.
Therefore, $(a, b, c)$ is a non-isosceles peculiar triangle.