Given $f(x)=x^{3}+m x^{2}+n x+5, m 、 n$ $\in \mathbf{Z}$. Find:
(1) All pairs $(m, n)$ such that $f(x)=0$ has 3 integer roots (including repeated roots);
(2) All pairs $(m, n)$ such that $f(x)=0$ has at least 1 integer root, and $0 \leqslant m \leqslant 5,0 \leqslant n \leqslant 5$.
(1) If $\alpha$ is an integer root of $f(x)=0$, then from $m, n \in \mathbf{Z}$ and $\alpha\left(\alpha^{2}+m \alpha+n\right)=-5$ we know $\alpha \mid 5$.

Let $\alpha, \beta, \gamma$ be the 3 integer roots of $f(x)=0$, then $\alpha \beta \gamma = -5$. Since 5 is a prime number, $\alpha, \beta, \gamma$ can only be -1, -1, -5 or 1, 1, -5 or 1, -1, 5, disregarding the order.

First case: $f(x)=(x+1)^{2}(x+5)=x^{3}+7 x^{2}+11 x+5, (m, n)=(7,11)$;

Second case: $f(x)=(x-1)^{2}(x+5)=x^{3}+3 x^{2}-9 x+5, (m, n)=(3,-9)$;

Third case: $f(x)=(x+1)(x-1)(x-5)=x^{3}-5 x^{2}-x+5, (m, n)=(-5,-1)$.
(2) Since $m \geqslant 0, n \geqslant 0$, $f(x)=0$ has no positive roots.
$f(x)=0$ has at least one integer root, which can only be -1 or -5.
Substituting $x=-1$ into the equation gives $m-n=-4$.
The integer solutions to this equation that satisfy $0 \leqslant m \leqslant 5, 0 \leqslant n \leqslant 5$ are only 2: $(m, n)=(1,5), (0,4)$.
Substituting $x=-5$ into the equation gives $5 m-n=24$.
The integer solution to this equation that satisfies $0 \leqslant m \leqslant 5, 0 \leqslant n \leqslant 5$ is only 1: $(m, n)=(5,1)$.