I’m not sure about the second question, but I think the first question can be answered as “Yes” with the help of semidirect products.
Let $\mathfrak{g}$ and $\mathfrak{k}$ be two Lie algebras and let $\theta$ be a homomorphism of Lie algebras from $\mathfrak{k}$ to ${\rm Der}(\mathfrak{g})$.
We can then form the semidirect product $\mathfrak{k} \ltimes_\theta \mathfrak{g} =: \mathfrak{h}$.
Its underlying vector space is given by the direct sum $\mathfrak{k} \oplus \mathfrak{g}$, and its Lie bracket is given by
$$
[ (x_1, y_1), (x_2, y_2) ]
=
( [x_1, x_2], \theta(x_1)(y_2) - \theta(x_2)(y_1) + [y_1,y_2] ) \,.
$$
The inclusion $i$ from $\mathfrak{k}$ to $\mathfrak{h}$ given by $i(x) = (x,0)$ is an injective homomorphism of Lie algebras, which identifies $\mathfrak{k}$ with a Lie subalgebra of $\mathfrak{h}$.
The inclusion $j$ from $\mathfrak{g}$ to $\mathfrak{h}$ given by $j(y) = (0,y)$ is also an injective homomorphism of Lie algebras, but it identifies $\mathfrak{g}$ with a Lie ideal of $\mathfrak{h}$.
The Lie bracket on $\mathfrak{h}$ is built in precisely such a way that
\begin{align*}
[i(x), j(y)]
&=
[ (x,0), (0,y) ]
\\
&=
( [x,0], \theta(x)(y) - \theta(0)(0) + [0,y] )
\\
&=
(0, \theta(x)(y))
\\
&=
j( \theta(x)(y) )
\end{align*}
for all $x \in \mathfrak{k}$, $y \in \mathfrak{g}$.
In other words, the restriction of the inner dervation $[i(x), -]$ to $\mathfrak{g}$ (when regarded a Lie ideal of $\mathfrak{h}$) is the derivation $\theta(x)$.
This constructions allows us to turn arbitrary derivations of $\mathfrak{g}$ into inner derivations on a suitable extension $\mathfrak{h}$ of $\mathfrak{g}$.
An extreme case of this is $\mathfrak{k} = \operatorname{Der}(\mathfrak{g})$ and $\theta = \mathrm{id}$.
The resulting semidirect product
$$
\mathfrak{h} := \operatorname{Der}(\mathfrak{g}) \ltimes_{\mathrm{id}} \mathfrak{g}
$$
is an extension of $\mathfrak{g}$, and every derivation of $\mathfrak{g}$ comes from an inner derivation of $\mathfrak{h}$.
More precisely, if $\delta$ is a derivation of $\mathfrak{g}$ then the restriction of the inner derivation $[(\delta, 0), -]$ of $\mathfrak{h}$ to $\mathfrak{g}$ is precisely $\delta$.
Yes. See Theorem 6.94 of Knapp, Lie groups, beyond an introduction.
This states: if $\mathfrak{g}$ is a real simple Lie algebra then either its complexification $\mathbf{C} \otimes_{\mathbf{R}} \mathfrak{g}$ is simple or its complexification is not simple, $\mathfrak{g}$ is a complex simple Lie algebra, and its complexification is isomorphic to $\mathfrak{g} \oplus \mathfrak{g}$. In either case, $\mathfrak{g}$ is a real form of a semi-simple Lie algebra.
Best Answer
Hint: We have $\mathfrak{gl}_n(\Bbb C)\cong \mathfrak{sl}_n(\Bbb C)\oplus \Bbb C$, where the first ideal is simple and the second one is abelian.
In general, if we have direct sum of ideals $L=L_1\oplus L_2$, we can express ${\rm Der}(L)$ by ${\rm Der}(L_1)$, ${\rm Der}(L_2)$ and the spaces ${\rm Hom}(L_1,Z(L_2))$, ${\rm Hom}(L_2,Z(L_1))$.
Since $L_1$ is semisimple, ${\rm Der}(L_1)={\rm Inn}(L_1)$ and $Z(L_1)=0$. Since $L_2$ is abelian, $Z(L_2)=L_2$ and ${\rm Der}(L_2)={\rm End}(L_2)$. For a reference see for example here. Note that for $L=\mathfrak{gl}_n(\Bbb C)$ the center $Z(L)=Z(L_2)$ is $1$-dimensional and is mapped into itself by every derivation.