2. $\frac{2 \sqrt{2}}{3}$.
From $1+3 t^{4} \geqslant 2 \sqrt{3 t^{4}}=\sqrt{12 t^{4}}$, we know $\sqrt{1+3 t^{4}} \geqslant \sqrt[4]{12} t$. Therefore,
$$
\frac{1}{\sqrt{1+3 t^{4}}}+\frac{t^{3}}{\sqrt[4]{12}} \geqslant \frac{1+t^{4}}{\sqrt{1+3 t^{4}}}=\frac{\sqrt{1+3 t^{4}}}{3}+\frac{2}{3 \sqrt{1+3 t^{4}}} \geqslant 2 \sqrt{\frac{2}{9}}=\frac{2 \sqrt{2}}{3},
$$

The equality holds if and only if $1+3 t^{4}=2$, i.e., $t=\frac{\sqrt[4]{27}}{3}$.
2. Let $t \in \mathbf{R}_{+}$, then the minimum value of $\frac{1}{\sqrt{1+3 t^{4}}}+\frac{t^{3}}{\sqrt[4]{12}}$ is