2. Inside the tetrahedron $ABCD$ there is a point $O$, satisfying $OA=OB=OC=4, OD=1$.
Find the maximum volume of the tetrahedron $ABCD$.
2. First, fix points $A$, $B$, $C$, and $O$. To maximize the volume of tetrahedron $ABCD$, the distance from point $D$ to plane $ABC$ should be maximized.

Since point $D$ moves on a sphere with center $O$ and radius 4, the volume is maximized when $OD \perp$ plane $ABC$.

Let the projection of $O$ on plane $ABC$ be point $E$, and $|OE| = x$. Then the distance from point $D$ to $\triangle ABC$ is $1 + x$.
And $EA = EB = EC = \sqrt{16 - x^2}$.
$\Rightarrow S_{\triangle ABC} \leq \frac{3 \sqrt{3}}{4}(16 - x^2)$.
(Here, we use: If $A$, $B$, $C$ are three points on a circle with radius $R$, then $S_{\triangle ABC} \leq \frac{3 \sqrt{3}}{4} R^2$.)
Therefore, $V_{\text{tetrahedron } ABCD} \leq \frac{\sqrt{3}}{4}(16 - x^2)(1 + x)$.
Consider the function $f(x) = (16 - x^2)(1 + x)$ for $x \in (0, 4)$.
It is easy to see that $f'(x) = -3x^2 - 2x + 16$.
Thus, $f(x)$ has a unique critical point $x = 2$ in the interval $(0, 3)$.
Hence, the maximum value of $f(x)$ in the interval $(0, 3)$ is $f(2) = 36$.

Therefore, the maximum value sought is $9 \sqrt{3}$.