3. Find all positive real numbers $t$ such that there exists an infinite set $X$ of real numbers, for which, for any $x, y, z \in X$ (where $x, y, z$ can be the same), and any real number $a$ and positive real number $d$, we have
$$
\max \{|x-(a-d)|,|y-a|,|z-(a+d)|\}>t d
$$
3. (1) $01$.
Let $x_{i}=\lambda^{i}, X=\left\{x_{1}, x_{2}, \cdots\right\}$.
Assume there exist $a \in \mathbf{R}, d \in \mathbf{R}$. and $x_{i}, x_{j}, x_{k}$ satisfying
$$
\begin{array}{l}
\max \left\{\left|x_{i}-(a-d)\right|,\left|x_{j}-a\right|,\left|x_{k}-(a+d)\right|\right\} \leqslant t d, \\
\left\{\begin{array}{l}
(-1-t) d \leqslant x_{i}-a \leqslant(-1+t) d, \\
-t d \leqslant x_{j}-a \leqslant t d, \\
(1-t) d \leqslant x_{k}-a \leqslant(1+t) d .
\end{array}\right.
\end{array}
$$
(2) - (1), (3) - (2) yields
$$
\left\{\begin{array}{l}
(1-2 t) d \leqslant x_{j}-x_{i} \leqslant(1+2 t) d, \\
(1-2 t) d \leqslant x_{k}-x_{j} \leqslant(1+2 t) d .
\end{array}\right.
$$

By $t<\frac{1}{2}$.
Thus, from (4) and (5) we get
$$
\begin{array}{l}
x_{j}-x_{i}>0, x_{k}-x_{j}>0 . \\
\text { Combining } \lambda>1 \text {, we know } i<j<k \text {, and } \\
\frac{x_{j}-x_{j-1}}{x_{j-1}}=\frac{x_{k}-x_{k-1}}{x_{k-1}} \\
=\lambda-1=\frac{1+2 t}{1-2 t},
\end{array}
$$

which is a contradiction.
Therefore, the infinite set $X$ constructed above meets the requirements.
(2) $t \geqslant \frac{1}{2}$.
We now prove: for any three elements $x<y<z$ in any infinite set $X$, there exist $a \in \mathbf{R}, d \in \mathbf{R}_{+}$, such that $\max \{|x-(a-d)|,|y-a|,|z-(a+d)|\} \leqslant t d$. In fact, let $a=\frac{x+z}{2}, d=z-x$. Then
$$
\begin{array}{l}
\max \{|x-(a-d)|,|y-a|,|z-(a+d)|\} \\
=\max \left\{\frac{z-x}{2},\left|y-\frac{x+z}{2}\right|, \frac{z-x}{2}\right\} \\
=\frac{z-x}{2}=\frac{1}{2} d \leqslant t d .
\end{array}
$$

In summary, $0<t<\frac{1}{2}$.