5. Given that $T$ is a moving point on the surface of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, and satisfies $|T A|=2|T B|$. Let the angle between $T D_{1}$ and the plane $A B C D$ be $\theta$. Then the maximum value of $\theta$ is $\qquad$
5. $\frac{\pi}{4}$.

As shown in Figure 2, with $B$ as the origin, the lines $B C$, $A B$, and $B B_{1}$ are taken as the $x$-axis, $y$-axis, and $z$-axis, respectively, to establish a spatial rectangular coordinate system.
Let the edge length of the cube be 2, and point $T(x, y, z)$.
Then $A(0,2,0)$.
Given $|T A|=2|T B|$
$$
\begin{aligned}
\Rightarrow & \sqrt{(x-0)^{2}+(y-2)^{2}+(z-0)^{2}} \\
& =2 \sqrt{x^{2}+y^{2}+z^{2}} \\
\Rightarrow & x^{2}+\left(y+\frac{2}{3}\right)^{2}+z^{2}=\frac{16}{9} .
\end{aligned}
$$

Thus, the trajectory of point $T$ is the intersection of the sphere with center $Q\left(0,-\frac{2}{3}, 0\right)$ and radius $\frac{4}{3}$ with the surface of the cube, which is the arcs $\overparen{E M G}$, $\overparen{G S F}$, and $\overparen{E N F}$ in Figure 2.

To maximize the angle between $T D_{1}$ and the base $A B C D$, the intersection point $R$ of $T D_{1}$ with the base $A B C D$ should be the closest to point $D$.
Therefore, point $T$ is on the arc $\overparen{E N F}$ and on $Q D$.
Thus, $D T=D Q-\frac{4}{3}=\frac{10}{3}-\frac{4}{3}=2$.
Hence, $\tan \theta=\frac{D D_{1}}{D T}=1$.

Therefore, the maximum value of $\theta$ is $\frac{\pi}{4}$.