Determine an equation of the line $g$ in coordinates $x', y'$ in terms of the basis $B$ 
Given is the equation $y = (2+\sqrt{3}) \cdot x \,\,$ which describes
  the line $g$. Also, the basis$B = \left\{\vec{b_1}; \vec{b_2}\right\} =
\left\{ \begin{pmatrix} \frac{\sqrt{3}}{2}\\  \frac{1}{2}
\end{pmatrix}; \begin{pmatrix}
-\frac{1}{2}\\  \frac{\sqrt{3}}{2} \end{pmatrix} \right\}$ is given.
Determine an equation of the line $g$ in coordinates $x', y'$ in terms
  of the basis $B$.

I'm not sure how this is supposed to work, so we have already an equation given and two coordinates we shall use to create another equation of line $g$ but in terms of the basis which is given.
Then I would insert the unknown coordinates into the current equation:
$y' = (2+\sqrt{3}) \cdot x'$
Then because it needs to be in terms of the basis $B$, I would do
$$y' = (2+\sqrt{3}) \cdot x' \begin{pmatrix} \frac{\sqrt{3}}{2}\\  \frac{1}{2}
\end{pmatrix}$$
But I don't know how to make use of the other basis vector and I think it's not good like that? : /

The straightforward way to approach this is to express $x$ and $y$ in terms of $x'$ and $y'$ and substitute into the equation. Recalling the definition of coordinates relative to a basis, you have $$\begin{pmatrix}x\\y\end{pmatrix} = x'\begin{pmatrix}{\sqrt3\over2}\\\frac12\end{pmatrix}+y'\begin{pmatrix}-\frac12\\{\sqrt3\over2}\end{pmatrix} = \begin{pmatrix}{\sqrt3\over2}x'-\frac12y' \\ \frac12x'+{\sqrt3\over2}y'\end{pmatrix}.$$ Substiting these expressions into the equation of the line produces $$\frac12x'+{\sqrt3\over2}y' = (2+\sqrt3)\left({\sqrt3\over2}x'-\frac12y'\right)$$ which I expect you’ll be able to rearrange into the form $y'=mx'+b$.  
More generally, you can rearrange the equation of any line through the origin into the form $ax+by=0$, which can be written in matrix form as $$\begin{pmatrix}a&b\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = 0.$$ If you have the coordinate transformation $(x,y)^T=B(x',y')^T$, substituting into this matrix equation produces $$\begin{pmatrix}a&b\end{pmatrix}B\begin{pmatrix}x'\\y'\end{pmatrix} = 0$$ so the coefficients of the equation of the line in the new basis are the components of the row vector $(a,b)B$.
