prove: max (w(E), w(E)) is a 1/2 approximation to the value OPT
I don't know the definition of $1/2$-approximation algorithm, but it looks like what you will need to do is the following argument:
For an enumeration $\sigma:V\to\{1,2,...,n\}$ of the vertices, let, for each input graph $G=(V,E)$, the output of the algorithm be $A_\sigma(G)=\max(w(\overrightarrow{E}),w(\overleftarrow{E}))$.
We have that $OPT(G)\leq w(E)$.
On the other hand,
$$
\begin{align}
A_\sigma(G)&amp;=\max(w(\overrightarrow{E}),w(\overleftarrow{E}))\\
&amp;\geq\frac{w(\overrightarrow{E})+w(\overleftarrow{E})}{2}\\
&amp;= \frac{w(E)}{2}
\end{align}$$
The last equation is because $\overrightarrow{E},\overleftarrow{E}$ is a partition of $E$.
Putting these two inequalities together we get $$OPT(G)\leq w(E)\leq 2A_\sigma(G)$$
Therefore $$\frac{A_\sigma(G)}{OPT(G)}\geq \frac{1}{2}$$