3. Let the areas of the four faces of a tetrahedron be $S_{1}, S_{2}, S_{3}, S_{4}$, and their maximum value be $S$. Let $\lambda=\frac{\sum_{i=1}^{4} S_{i}}{S}$. Then $\lambda$ must satisfy
(A) $2<\lambda \leqslant 4$;
(B) $3<\lambda<4$;
(C) $2.5<\lambda \leqslant 4.5$;
(D) $3.5<\lambda<5.5$.
3. (A)

Since
$$
S_{i} \leqslant S \quad(i=1,2,3,4),
$$

we have

Furthermore,
$$
\begin{array}{c}
\sum_{i=1}^{4} S_{i} \leqslant 4 S . \\
\lambda=\frac{\sum_{i=1}^{4} S_{i}}{S} \leqslant 4 .
\end{array}
$$

In particular, when the tetrahedron is a regular tetrahedron, the equality in (1) holds, thus negating (B).

Consider any regular triangular pyramid where the dihedral angles between the lateral faces and the base are all $45^{\circ}$. Let $S_{4}$ be the area of the base of this regular triangular pyramid, then
$$
S=S_{4}=\left(S_{1}+S_{2}+S_{3}\right) \cos 45^{\circ}=\frac{\sqrt{2}}{2}\left(S_{1}+S_{2}+S_{3}\right),
$$

Thus,
$$
S_{1}+S_{2}+S_{3}=\sqrt{2} S,
$$
$$
S_{1}+S_{2}+S_{3}+S_{4}=(1+\sqrt{2}) S \text {. }
$$

In this case, $\lambda=\frac{\sum_{i=1}^{4} S_{i}}{S}=1+\sqrt{2}<2.5$,
thus negating (C) and (D). Therefore, the answer is (A).