Let $Ad:Pin(V,q)\to O(V,q)$ be defined as the following where $q$ is a non-degenerate quadratic form over a finite dimensional vector space $V$. Let $a:V\to V$ be $v\to -v$ map and induce $a:Cl(V,q)\to Cl(V,q)$ where $CL(V,q)$ is clifford algebra generated by $V,q$. $Pin(V,q)=<v\in V\vert q(v)=\pm 1>$ where $<->$ means group generated by $-$. Then $Ad:Pin(V,q)\to O(V,q)$ is defined by $v\in V$, $x\in Pin(V,q)\to a(x)\cdot v\cdot x^{-1}$. If $V$ is over $R$ or $C$, then $Ad$ is clearly surjection. Now one can restrict the map further to $Spin(V,q)$ with all elements made of even number of products of elements of $V$.(i.e. $v_1,\dots, v_{2k}\in V, q(v_i)\neq 0$ where products like $v_1\cdot\dots\cdot v_{2k}$ generates $Spin(V,q)$.)
"In the cases where $Ad$ is not surjective, we still have the following fact, which is interesting because the group $SO(V,q)$ is often almost a simple group."
$\textbf{Q1:}$ Why this is interesting? What is the author pointing to here?
$\textbf{Q2:}$ "$SO(V,q)$ is often almost a simple group". What does this even mean? I knew $Ad(Pin(V,q)),Ad(Spin(V,q))$ are normal subgroups of $O(V,q)$. $SO(V,q)$ is normal subgroup of $O(V,q)$ by determinant map. Clearly $Ad(Spin(V,q))$ has image in $SO(V,q)$. Thus $Ad(Spin(V,q))$ is already normal in $SO(V,q)$ as well. This does not say whether $Ad(Spin(V,q))$ has normal subgroup.
Ref. Spin Geometry, Lawson Chpt 1, Sec 2.