3. Let $\Gamma$ be a circle with radius 1. A set of triangles $T$ is called "good" if and only if it satisfies the following conditions:
(1) Each triangle in $T$ is an inscribed triangle in circle $\Gamma$;
(2) No two triangles in $T$ have a common interior point.

Find all positive real numbers $t$ such that for each positive integer $n$, there exists a good set of $n$ triangles, and the perimeter of each triangle is greater than $t$.
3. $t \in(0,4]$.

First, prove: there exists a good set composed of $n$ triangles, each with a perimeter greater than 4.
This proves that for all $t \leq 4$, the conclusion holds.
Next, use a recursive method to construct a convex $n+2$-gon $B A_{1} A_{2} \cdots A_{n} C$ inscribed in a circle $\Gamma$, such that $B C$ is the diameter of circle $\Gamma$, and the $n$ triangles:
$$
\triangle B A_{1} A_{2}, \triangle B A_{2} A_{3}, \cdots, \triangle B A_{n-1} A_{n}, \triangle B A_{n} C
$$
form a good set.
When $n=1$, any inscribed $\triangle B A_{1} C$ in circle $\Gamma$ satisfies the condition, where $B C$ is the diameter of circle $\Gamma$. This is because the perimeter of $\triangle B A_{1} C$ is greater than $2 B C=4$.
Assume the convex $n+2$-gon $B A_{1} A_{2} \cdots A_{n} C$ satisfies the condition.
Since $A_{n} B + A_{n} C + B C > 4$, we can choose a point $A_{n+1}$ sufficiently close to point $C$ on the minor arc $\overparen{C A}_{n}$, such that $A_{n} B + A_{n} A_{n+1} + B A_{n+1} > 4$ still holds. Thus, the new triangles $\triangle B A_{n} A_{n+1}$ and $\triangle B A_{n+1} C$ both have perimeters greater than 4. Therefore, the proof by mathematical induction is completed (Figure 5).
Next, prove: there does not exist $t>4$ that satisfies the condition.
Assume there exists a good set $T$ containing $n\left(n \in \mathbf{Z}_{+}\right)$ triangles, each with a perimeter greater than $t$.
Let $t=4+2 \varepsilon(\varepsilon>0)$.
Lemma There exists a positive constant $\sigma=\sigma(\varepsilon)$, such that for any triangle $\Gamma_{0}$ inscribed in circle $\Gamma$ with perimeter $2 p \geqslant 4+2 \varepsilon$, its area $S\left(\Gamma_{0}\right) \geqslant \sigma$.

Proof Let the side lengths of triangle $\Gamma_{0}$ be $a, b, c$.

Since triangle $\Gamma_{0}$ is inscribed in circle $\Gamma$, $a, b, c$ are all no more than 2.
Thus, $p-a \geqslant 2+\varepsilon-2=\varepsilon$.
Similarly, $p-b \geqslant \varepsilon, p-c \geqslant \varepsilon$.
By Heron's formula,
$$
\begin{array}{l}
S\left(\Gamma_{0}\right)=\sqrt{p(p-a)(p-b)(p-c)} \\
\geqslant \sqrt{(2+\varepsilon) \varepsilon^{3}} .
\end{array}
$$

Let $\sigma(\varepsilon)=\sqrt{(2+\varepsilon) \varepsilon^{3}}$, and the lemma is proved.
By the lemma, the total area $S$ of all triangles in set $T$ is at least $n \sigma(\varepsilon)$.

Since $S$ does not exceed the area of the disk bounded by circle $\Gamma$, which is $\pi$, we have $n \sigma(\varepsilon) \leqslant S \leqslant \pi$.
This implies that $n$ is bounded, leading to a contradiction.
In conclusion, $t \in(0,4]$.