Given Lines are,
$$
\begin{aligned}
& y=3 x+1 \Rightarrow m_1=3 \Rightarrow 2 y=x+3 \\
& y=\frac{1}{2} x+\frac{3}{2} \Rightarrow m_2=\frac{1}{2} \\
& y=m x+4
\end{aligned}
$$
According to the question,
$$
\left|\frac{m_1-m}{1+m_1 m}\right|=\left|\frac{m_2-m}{1+m_2 m}\right|
$$
$\left\{\because\right.$ Angle between lines is $\left.\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\right\}$
$$
\begin{aligned}
& \left|\frac{3-m}{1+3 m}\right|=\left|\frac{\frac{1}{2}-m}{1+\frac{1}{2} m}\right| \\
& \left|\frac{3-m}{1+3 m}\right|=\left|\frac{1-2 m}{2+m}\right| \\
& \frac{3-m}{1+3 m}= \pm\left|\frac{1-2 m}{2+m}\right|
\end{aligned}
$$
Case (i)
$$
\begin{aligned}
\frac{3-m}{1+3 m} & =-\left(\frac{1-2 m}{2+m}\right) \\
(3-m)(2+m) & =(2 m-1)(1+3 m) \\
6+m-m^2 & =6 m^2-m-1 \\
7 m^2-2 m-7 & =0 \\
m & =\frac{2 \pm \sqrt{(-2)^2-4 \cdot 7(-7)}}{2 \cdot 7} \\
& =\frac{2 \pm \sqrt{4+196}}{14}=\frac{2 \pm 10 \sqrt{2}}{14}=\frac{1 \pm 5 \sqrt{2}}{7} \\
\therefore \quad m & =\frac{1 \pm 5 \sqrt{2}}{7}
\end{aligned}
$$
Hence, option (4) is correct.