I have been thinking about this the past few days in preparation for an exam at EPFL as a result of some really shitty course notes. My familiarity with the subject is thus rather poor but at least I sympathize with your plight for clarity.
1 . I think that key to working with this problem is to first make concrete what the complexification of $\mathfrak{su}(2)$, $\mathfrak{su}(2)_\mathbb{C}$, really is and what its algebra is. We know that the natural basis of the $\mathfrak{su}(2)$ are the Pauli matrices $\{\sigma_1, \sigma_2, \sigma_3\}$ with the familiar Lie Bracket $[\sigma_i, \sigma_j] = i \varepsilon_{ijk}\sigma_k$. This is a REAL vector space and the complexification is a particular complex vector space where the Lie bracket is essentially what we expect it to be when treating the bracket as if it is linear over $i$ as well
$\mathfrak{su}(2)_\mathbb{C}$ is the Lie algebra of formal sums $u + iv$ where $u,v \in \mathfrak{su}(2)$ and where the complexified Lie-bracket expressed in terms of the real Lie bracket is
$$[x + iy, u + iv]_{\mathbb{C}} = ([x,u] - [y,v]) + i([x,v] + [y,u])$$
I wont write the complex sign again as its easy to take as implicit. Now that we hopefully agree on the definition I am probably going to annoy you by viewing complexified algebras as real algebras of twice the dimension because I find this situation to be more transparent. I am free t view my complexified algbra as a real algebra and in this picture the most natural basis we can come up with is
$$\sigma_1, \sigma_2, \sigma_3, i \sigma_1, i\sigma_2, i\sigma_3$$
I check the resulting Lie brackets and we end up with
$$[\sigma_i, \sigma_j] = i \varepsilon_{ijk}\sigma_k \\ [\sigma_i, i\sigma_j] = i \varepsilon_{ijk}(i\sigma_k) \\ [i\sigma_i, i \sigma_j ] = -i \varepsilon_{ijk}\sigma_k$$
We easily see a correspondence
$$J_j \leftrightarrow \sigma_j \qquad K_j\leftrightarrow i\sigma_j$$
and conclude
$$\mathfrak{so}(1,3) \simeq \mathfrak{su}(2)_\mathbb{C}$$
thus to be it looks like it is the REAL $\mathfrak{so}(1,3)$ which is isomorphic to the complexification of $\mathfrak{su}(2)$ (but also viewed as a REAL Lie algbera, of real dimension $6$). I find this to be a much more transparent way of arriving at the isomorphism rather than going via the complexification.
2. To me this looks like it will imply
$$\mathfrak{so}(1,3)_\mathbb{C} \simeq (\mathfrak{su}(2)_\mathbb{C})_\mathbb{C} \simeq \mathfrak{su}(2)_\mathbb{C} \oplus_\mathbb{C}\mathfrak{su}(2)_\mathbb{C} $$
I have to admit I don't know how to make sense of going via the complexification of $\mathfrak{so}(1,3)$ neither. I had an argument planned out but it collapsed and I reverted to the one above. Maby I'll try to fix this if you come back and discuss it with me.
3. I started thinking about this but I think you actually mean $\mathfrak{sl}(2,\mathbb{R})_\mathbb{C} \simeq \mathfrak{sl}(2,\mathbb{C}) \simeq \mathfrak{su}(2)_\mathbb{C}$? $\mathfrak{sl}(2,\mathbb{C})$ is a real vector space made up of traceless complex matrices so the 6 most obvious basis matrices are
$$\alpha_1 = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}, \alpha_2 = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}, \alpha_3 = \begin{pmatrix}0& 0 \\ 1 & 0\end{pmatrix}, \; \text{and} \; i\alpha_1, i\alpha_2, i\alpha_3$$
From this we can find an explicit change of basis to the complexified Pauli matrices $$\sigma_1 = \alpha_2 + \alpha_3, \quad \sigma_2 = i\alpha_1 - i\alpha_3, \quad \sigma_3 = \alpha_1\\
i \sigma_1 = i\alpha_2 + i\alpha_3, \quad i\sigma_2 = \alpha_1 - \alpha_3, \quad i\sigma_3 = i\alpha_1$$
and since the the bracket is the commutator we see that the Lie-structures of these two Lie algebras are the same meaning they are the same.
4. To me it looks like we will have $\mathfrak{so}(1,3) \simeq \mathfrak{sl}(2,\mathbb{C})$ (where the latter is viewed as a $6$-dimensional real Lie algbera) which kind of surprises me.
5. Well if 4. holds then it should hold.
An irreducible representation $\phi:\mathfrak{g}_\mathbb{C} \rightarrow GL(\mathbb{C,n})$ is a Lie algebra homomorphism by definition. That is, it's linear.
So, to consider your example, we see that for $x,y \in \mathfrak{so}(3)$,
\begin{align}
\pi(x+iy) = \pi(x)+i\pi(y)
\end{align}
which shows that working with
\begin{align}
\pi: \mathfrak{so}(3) \rightarrow GL(\mathbb{C,3})
\end{align}
is the same as working with
\begin{align}
\pi: \mathfrak{so}(3)_\mathbb{C} \rightarrow GL(\mathbb{C,3})
\end{align}
(if this doesn't convince you, try to construct a counter example to the proposition and you'll see that it's impossible precisely because any irreducible representation is a homomorphism).
Best Answer
As regards question 2, you seem to be using subscripts for two different things. If subscript $\mathbb C$ stands for complexification but subscript $\mathbb R$ stands for scalar restriction to $\mathbb R$, then a) please use a different notation and b) it is still not true. Cf. answers to Are Lie algebra complexifications $\mathfrak g_{\mathbb C}$ equivalent to Lie algebra structures on $\mathfrak g\oplus \mathfrak g$? and Is the complexification of $\mathfrak{sl}(n, \mathbb{C})$ itself?.
What is true is
$$sl(2, \mathbb C) \otimes_{\mathbb R} \mathbb C\simeq sl(2, \mathbb C) \oplus sl(2, \mathbb C)$$
where both sides are complex Lie algebras and the isomorphism is one of complex LAs. If you view both sides as real Lie algebras via scalar restriction (and we denote that by the subscript $\mathbb R$), then this implies, of course, that there is an isomorphism of real Lie algebras
$$(sl(2, \mathbb C) \otimes_{\mathbb R} \mathbb C)_\mathbb R \simeq sl(2, \mathbb C)_{\mathbb R} \oplus sl(2, \mathbb C)_{\mathbb R}.$$
[So the left hand side is the complexification but viewed as a real algebra again, i.e. complexification followed by scalar restriction; if you insist on sticking to your confusing notation -- which you shouldn't -- you should write the left hand side as $(sl(2, \mathbb C)_{\mathbb C})_{\mathbb R}$.]