For 1, there exists $\psi\in \operatorname{Aut}(S_6)\setminus \operatorname{Inn}(S_6)$ s.t. $\psi^2=\text{id}$.
$\quad\psi:(12)\mapsto(15)(23)(46), (13)\mapsto(14)(26)(35), (14)\mapsto(13)(24)(56),\\\qquad (15)\mapsto(12)(36)(45), (16)\mapsto(16)(25)(34).$
Therefore $\operatorname{Aut}(S_6)\cong S_6\rtimes\mathbb Z_2$.
For 2, we have short exact sequence for groups: $1\to S_6\overset{f}{\to}\operatorname{Aut}(S_6)\overset{\pi}{\to} \mathbb Z_2\to 1 $, $\mathbb Z_2=\{\pm1,\times\}$.
This sequence right splits, so there exists homomorphism $g:\mathbb Z_2 \to \operatorname{Aut}(S_6)$ s.t. $\pi\circ g=\text{id}.$
Let $g(-1)=\psi\not\in \operatorname{Inn}(S_6)$, then $g(1)=\psi^2=\text{id}$.
$f:S_6\to \operatorname{Inn}(S_6)$, $g:\mathbb Z_2 \to \langle\psi\rangle$.
Claim: $\langle\psi\rangle$ is not normal subgroup of $\operatorname{Aut}(S_6)$, so $\operatorname{Aut}(S_6)\not \cong S_6\times\mathbb Z_2$.
For $\sigma\in S_6$, define $\gamma_\sigma \in \operatorname{Inn}(S_6)$ to be action by conjugation of $\sigma$.
It's sufficient to prove $\gamma_\sigma\psi\gamma_\sigma^{-1}\neq\psi$, i.e.$\gamma_\sigma\psi\neq\psi\gamma_\sigma$ for some $\sigma\in S_6$.
Let $\sigma=(12)$, $\gamma_\sigma\psi((12))=\gamma_\sigma((15)(23)(46))=(12)(15)(23)(46)(12)=(13)(25)(46)$.
$\psi\gamma_\sigma(12)=\psi((12))=(15)(23)(46)$. $\gamma_\sigma\psi\neq\psi\gamma_\sigma$ for $\sigma=(12)$.
Thus $\operatorname{Aut}(S_6)\cong S_6\rtimes\mathbb Z_2$ and $\operatorname{Aut}(S_6)\not \cong S_6\times\mathbb Z_2$.
For 3, fix $1\neq\alpha\in A_n$, $c_\alpha\in\text{Inn}(A_n)$ is action by conjugation of $\alpha$.
Define $\varphi:\text{Aut}(S_n)\to\text{Aut}(A_n)$, $\varphi(\beta)=\beta c_\alpha \beta^{-1}$ for $\beta\in \text{Aut}(S_n)$.
Easy to check $\varphi$ is monomorphism, so $\text{Aut}(S_n)\leqslant\text{Aut}(A_n)$
Together with $[\text{Aut}(A_6):\text{Inn}(S_n)]\leqslant2$ and $[\text{Aut}(S_6):\text{Inn}(S_n)]=2$, we have
$\text{Aut}(A_6)=\text{Aut}(S_6)$.
Best Answer
Claim:
Short exact sequence $1 \longrightarrow \operatorname{Inn}\left(A_{6}\right) \longrightarrow \operatorname{Aut}\left(A_{6}\right) \longrightarrow \operatorname{Out}\left(A_{6}\right) \longrightarrow 1$ is not right split,
where $\operatorname{Inn}\left(A_{6}\right)\cong A_6$, $\operatorname{Aut}\left(A_{6}\right)\cong\operatorname{Aut}\left(S_{6}\right)\cong S_6\rtimes \mathbb Z_2$ and $\operatorname{Out}\left(A_{6}\right)\cong\mathbb Z_2\times \mathbb Z_2$.
Proof:
$1$. Prerequisites:
(1) Element in $\operatorname{Aut}\left(A_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$ swaps conjugate classes $(abc)$ and $(abc)(def)$ in $A_6$.
(2) Element in $\operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$ swaps conjugate classes $(ab)$ and $(ab)(cd)(ef)$ in $S_6$,
swaps conjugate classes $(abc)$ and $(abc)(def)$ in $A_6$(also in $S_6$).
$2$. Suppose the sequence right splits and $\operatorname{Out}\left(A_{6}\right)\cong\mathbb Z_2\times \mathbb Z_2\cong\langle \sigma\rangle\langle\rho\rangle\leqslant\operatorname{Aut}\left(A_{6}\right)$
where $\sigma, \rho\in \operatorname{Aut}\left(A_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$, then $\langle \sigma\rangle\langle\rho\rangle\cap\operatorname{Inn}(A_6)=1$.
Since $\operatorname{Aut}\left(A_{6}\right)=\operatorname{Aut}\left(S_{6}\right)$, $\rho$ and $\sigma$ can be considered as elements in $\operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$.
If $\sigma, \rho\in \operatorname{Inn}\left(S_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$, then $\sigma\rho\in \operatorname{Inn}\left(A_{6}\right)$. Contradiction.
If $\rho\in \operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$ and $\sigma\in \operatorname{Inn}\left(S_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$, then $\sigma\rho\in \operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$.
$\operatorname{Out}\left(A_{6}\right)\cong\mathbb Z_2\times \mathbb Z_2\cong\langle \sigma\rangle\langle\rho\rangle\cong\langle \sigma\rho\rangle\langle\rho\rangle$ and $\sigma\rho, \rho\in \operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$.
So if the sequence right splits, we can always assume
$\operatorname{Out}\left(A_{6}\right)\cong\mathbb Z_2\times \mathbb Z_2\cong\langle \sigma\rangle\langle\rho\rangle\leqslant\operatorname{Aut}\left(A_{6}\right)$ where $\rho, \sigma\in\operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$.
$3$. $[\operatorname{Aut}\left(S_{6}\right):\operatorname{Inn}\left(S_{6}\right)]=2$ , $\sigma\operatorname{Inn}\left(S_{6}\right)=\rho\operatorname{Inn}\left(S_{6}\right)$, $\rho^{-1}\sigma\in\operatorname{Inn}(S_6)$.
Suppose $\rho^{-1}\sigma=c_\gamma$, where $c_\gamma$ is action of conjugation by $\gamma\in S_6$.
Since $\langle \sigma\rangle\langle\rho\rangle\cap\operatorname{Inn}(A_6)=1$, $\gamma\in S_6\setminus A_6$ is odd permutation.
$(\rho^{-1}\sigma)^2=c_\gamma^2=1$ gives $\gamma^2=1$, $\gamma$ is transposition or product of three disjoint transpositions.
$\sigma\rho=\rho\sigma$ gives $\rho(\gamma)=\gamma$.
But $\rho\in\operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$ swaps conjugate classes $(ab)$ and $(ab)(cd)(ef)$.
Contradiction. $\Box$