Georgi-Glashow model and the VEV of the scalar field Consider the Georgi-Glashow model, an $SU(2)$ gauge theory with a real scalar in the adjoint (thus a 3-vector in the colour space) $\phi$. The Lagrangian is 
$$ L = -\frac{1}{4g^2} F_{\mu \nu}^{\, a} F^{\mu \nu \, a}+ \frac{1}{2}(D_{\mu}\phi^a)(D^{\mu}\phi^a) - \lambda(\phi^a \phi^a - v^2)^2 $$
where everything is as usual regarding the field strength and the covariant derivative. Also, let us write the scalar field in matrix form as
$$\phi = \frac{1}{2}\phi^a \sigma^a$$
where $\sigma^a$ are the Pauli matrices as usual. 
Now we can see already that the scalar field is self-interacting, right? Therefore it is forced to develop a vacuum expectation value (VEV). I am interested to see what this VEV is and how $SU(2)$ is broken down to $U(1)$. 
In my notes it says that the VEV developed is the following one
$$ \phi_{vac.}^a = v\delta^{3a}, \,\,\,\,\, \phi_{vac.} = v\frac{\sigma^3}{2}$$
because $\phi_{vac.}=\phi_{vac.}^a\sigma^3/2 = v\sigma_{3}/2$. Now, I understand why we can always bring the VEV into this form (by a global color rotation).
Question 1. A color rotation in the third (colour) axis will not change the VEV:
$$ e^{ia\sigma^3} \phi_{vac.} e^{-ia\sigma^3} = \phi_{vac.} $$
But why will the other rotations, i.e. $$ e^{ia\sigma^1} \phi_{vac.} e^{-ia\sigma^1}$$  change it? I do not understand this point. Does it have to do something with the fact that different Pauli matrices do not commute? 
Question 2. Why the 1st and 2nd components of the gauge field, i.e. $A_{\mu}^1$ and $A_{\mu}^2$ form the W-bosons and why they are massive? Why are they defined as 
$$ W_{\mu}^{\pm} = \frac{1}{\sqrt{2}g}(A_{\mu}^1 + iA_{\mu}^2)?$$
Question 3. What changes if we choose a different representation for the scalar field? I know we can choose the spinor representation for $\phi$. So what will be different in the end? What about other representations, e.g. the fundamental? What are the fundamental differences to the SM Higgs model? There for example the Higgs is a complex scalar, right?
 A: The scalar potential of your theory is
$$V(\phi) = \lambda (\phi^a \phi^a - v^2)^2,$$
where I suspect you meant to take the square as I've written here. This potential is minimized when $\sqrt{\phi^a \phi^a} = v$. Think of $\phi=\frac{1}{2} \phi^a \sigma^a$ as a vector with components $\phi^a$ in a 3-dimensional vector space with basis vectors $\sigma_a/2$. The equation $\sqrt{\phi^a \phi^a}=v$ says that the norm of this vector is $v$ at the minima of the potential. The minimal locus of the potential is therefore a 2-sphere of radius $v$ in this 3d space, consisting of vectors whose norm is fixed to $v$. 
Choose at random one of these minimal field configurations, say $\phi_0=\frac{1}{2} v \sigma^3$ (the "north pole" of the 2-sphere, if you like). Consider how a gauge transformation acts on this field configuration. Since the scalar transforms in the adjoint representation of the gauge group, a gauge transformation $e^{iT} \in \mathrm{SU}(2)$ acts on $\phi_0$ as $\phi_0 \mapsto e^{i T} \phi_0 e^{-iT}$, or, infinitesimally, $\delta \phi_0 = i [T,\phi_0]$. Here, $T = \frac{1}{2} T^a \sigma^a$ is an arbitrary element of $\mathrm{su}(2)$, the Lie algebra of $\mathrm{SU}(2)$. Our field configuration $\phi_0 = \frac{1}{2} v \sigma^3$ is therefore invariant under this gauge transformation when $T \propto \sigma_3$, while $\phi_0$ is not invariant if $T$ has support along $\sigma^1$ or $\sigma^2$. We find that a single generator of $\mathrm{su(2)}$ (which generates a $\mathrm{U(1)}$ subgroup of $\mathrm{SU(2)}$) leaves $\phi_0$ invariant, while the other two generators act non-trivially. 
In such a situation, we say the $\mathrm{SU(2)}$ gauge symmetry has been Higgsed to a $\mathrm{U(1)}$ subgroup. To determine the spectrum of fields in the Higgsed theory, perform a field redefinition $\phi' = \phi- \phi_0$, such that the potential is now minimized for $\phi' = 0$. If you replace $\phi$ by $\phi'+\phi_0$ in your Lagrangian and expand it, you will find that the gauge fields $A_1$ and $A_2$ have become massive (in unitary gauge), while $A_3$ remains massless. You can immediately determine the gauge field masses by simply replacing $\phi$ by $\phi_0$ in the scalar kinetic term $(D_\mu \phi^a) (D^\mu \phi^a)$:
\begin{align}(D_\mu \phi_0^a)(D^\mu \phi_0^a) =& (g \epsilon^{abc} A_\mu^b v \delta^{c,3})(g \epsilon^{ade} A_\mu^d v \delta^{e,3})\\
=& g^2 v^2 \epsilon^{ab3} \epsilon^{ad3} A_\mu^b A_\mu^d\\
=&g^2 v^2 (A_\mu^1 A_\mu^1+A_\mu^2A_\mu^2).
\end{align}
Thus, $A_1$ and $A_2$ have each acquired a mass of order $gv$. They are often traded for the "complex gauge fields" $W_\pm = A_1 \pm i A_2$ because it is $W_\pm$ which appears in cubic interactions with matter. I will leave the explicit expansion of the Lagrangian for you to work out.
That should resolve your first two questions. Your third question asks what will happen if you take the scalar to live in a different representation. The analysis proceeds in the same way, so let me summarize the procedure for an arbitrary gauge group and representation. One begins with a scalar potential $V(\phi)$ and determines the fields configurations which minimize it, $\mathcal{M}_0 = \left \{\phi_0: V'(\phi_0) = 0, V''(\phi_0) > 0 \right\}$. Suppose the action is invariant under a symmetry group $G$, and that $\phi$ belongs to a linear representation $R$ of $G$. In other words, $\phi$ transforms as $\phi \mapsto R(g) \phi$, where $R(g)$ is the matrix representation of $g$. For an infinitesimal transformation (when $G$ is continuous), $\delta \phi = i T \phi$, where $R(g)=e^{iT}$. 
Since the action is invariant under $G$, $V(\phi_0) = V(R(g) \phi_0)$ for any $g \in G$. 
Thus $G$ maps $\mathcal{M}_0$ to itself. $R(g)$ need not leave $\phi_0$ invariant, however. In general, only a subgroup $H \subset G$ will leave $\phi_0$ invariant. That is, among the list of generators $\{T_a\}$ of $G$, some subset (the "unbroken" generators) will leave $\phi_0$ invariant, $\delta \phi_0 = i T_a \phi = 0$, while the remainder will not (the "broken" generators). The unbroken generators generate the subgroup $H$, while the broken generators correspond to the coset $G/H$. 
If $G$ is a global symmetry, we say that it has been spontaneously broken to the subgroup $H$. If $G$ is a gauge symmetry, we say that it has been Higgsed to $H$. The gauge fields along the generators of $H$ remain massless, while the gauge fields along the broken generators become massive (again, in unitary gauge). 
To ensure you understand this procedure, you should carry out this line of analysis for various other examples. I've shown you how the analysis goes for an $\mathrm{SU(2)}$ gauge theory Higgsed by an adjoint. You can come up with other examples of gauge groups and representations and work out the details, or study the many examples presented in the many field theory books out there.
By the way, the Georgi-Glashow model refers to an $\mathrm{SU(5)}$ gauge theory, not an $\mathrm{SU(2)}$ gauge theory. 
