[Solution] Consider $|f(x)+4|=| 3 x+2+4|=3| x+2 |$. If $|f(x)+4|=3|x+2|<a$, it is sufficient that $|x+2|<\frac{a}{3}$. Therefore, whenever $|x+2|<b, \quad b \leqslant \frac{a}{3}$, we always have $|f(x)+4|<a$. Hence, the answer is $(A)$.
4.50 If for all real numbers $x, f(x)=3 x+2$, then the statement “ $|f(x)+4|<a, \quad b>0$ ” is true under the condition
(A) $b \leqslant \frac{a}{3}$.
(B) $b>\frac{a}{3}$.
(C) $a \leqslant \frac{b}{3}$.
(D) $a>\frac{b}{3}$.
(E) This statement is never true.
(25th American High School Mathematics Examination, 1974)