We can define the projective $U(N)$ and $SU(N)$ as $$PSU(n)=\frac{SU(n)}{Z(SU(n))}=\frac{SU(n)}{Z_n}$$ and $$PU(n)=\frac{U(n)}{Z(U(n))}=\frac{U(n)}{U(1)}.$$ Here $Z(G)$ means the center of $G$. And it turns out that $$PSU(n)=PU(n)=\frac{U(n)}{Z(U(n))}=\frac{U(n)}{U(1)}=\frac{SU(n)}{Z(SU(n))}=\frac{SU(n)}{Z_n}.$$
Do we define the projective Spin group and Spin$^C$ group as follows: $$PSpin(n)=\frac{Spin(n)}{Z(Spin(n))}=?$$ and $$PSpin^C(n)=\frac{Spin^C(n)}{Z(Spin^C(n))}=?$$ If so, what precisely these groups are for any $n$?
Here $ {Spin}(n)/\mathbb{Z}_2= {SO} (n)$ and $ {Spin} ^{\mathbf {C} }(n)/\mathbb{Z}_2= {SO} (n)\times \operatorname {U} (1)$.
A good place to begin, at least for the spin groups, would be the wikipedia page which answers your first question in detail. Recall that the centre of the spinor groups is given by
$Z(Spin_n)=\begin{cases}\mathbb{Z}_2,&n\equiv1,3\,(4)\\ \mathbb{Z}_4&n\equiv 2\,(4)\\ \mathbb{Z}_2\oplus\mathbb{Z}_2&n\equiv0\;(4)\end{cases}$
Quotienting by the whole centre gives you something isomorphic to $PSO(n)$. Quotienting by a proper subgroup (in the second two cases) gives you a semi-spinor group. This is explained in more detail on wiki.
As for $Spin^c_n$, its centre is $Z(Spin_n)\times_{\mathbb{Z}_2}S^1$, and quotienting by this gives you something isomorphic to $Spin_n/Z(Spin_n)$. That is, something isomorphic to $PSO(n)$.