\[3x^{2} + bx + 4 = 0;\ \ \ \ \ x_{1} = 4\]
\[\left\{ \begin{matrix} x_{1} + x_{2} = - \frac{b}{3}\ \ (1) \\ x_{1} \cdot x_{2} = \frac{4}{3}\ \ \ (2)\ \ \ \\ \end{matrix} \right.\ \]
\[(1)\ 4 \cdot x_{2} = \frac{4}{3}\]
\[x_{2} = \frac{4}{3}\ :4\]
\[x_{2} = \frac{1}{3}.\]
\[(2)\ 4 + \frac{1}{3} = - \frac{b}{3}\]
\[\frac{13}{3} = - \frac{b}{3}\]
\[b = \frac{13}{3} \cdot ( - 3) = - 13.\]
\[Ответ:\ b = - 13;\ \ x_{2} = \frac{1}{3}.\]