You can work out the answers to these questions using the material in Chapter 11 of the book Spinors and Calibrations by F. Reese Harvey. You will also need to recall that, for $N\not=4$, the group of automorphisms of $\mathrm{Spin}(2N)$ is $\mathrm{O}(2N)$ rather than $\mathrm{SO}(2N)$. The answers depend to some extent on $N$ modulo $4$.
The point is that $\mathrm{Spin}(2N)$ is represented in $\mathbb{C}^{2^{N-1}}$ as either the full spinor space (when $N$ is odd) or a semi-spinor space (when $N$ is even), and the commuting ring of this representation (which may include complex conjugation) depends on $N$ modulo $4$.
For example, when $N=4$, the representation of $\mathrm{Spin}(8)$ is actually the complexification of a 8-dimensional real representation of $\mathrm{Spin}(8)$, whose image in $\mathrm{O}(8)$ is just a double cover onto $\mathrm{SO}(8)$. (In particular, this semi-spinor representation is not faithful.) Thus, for clarity, let me call the image of this representation $\mathrm{SO}'(8)\subset\mathrm{O}(8)\subset\mathrm{U}(8)$. The normalizer of $\mathrm{SO}'(8)$ in $\mathrm{O}(8)$ is $\mathrm{O}(8)$, and conjugation by elements of $\mathrm{O}(8)$ induce all of the automorphisms of $\mathrm{SO}'(8)\simeq \mathrm{SO}(8)$. Noting that the centralizer of $\mathrm{SO}'(8)$ in $\mathrm{U}(8)$ is simply $S^1 = \{\lambda I_8 \ |\ |\lambda|^2=1\ \}$, it now follows easily that the normalizer of $\mathrm{SO}'(8)$ in $\mathrm{U}(8)$ is simply $S^1{\cdot}\mathrm{O}(8)$, which has two components.
Meanwhile, when $N=5$, the group $\mathrm{Spin}(10)$ embeds into $\mathrm{SU}(16)$ and this irreducible representation on $\mathbb{C}^{16}$ is not the complexification of a 16-dimensional real representation. Moreover, the conjugate representation on $\mathbb{C}^{16}$ is not isomorphic (as a complex representation) to the given representation. In particular, an element of the normalizer of $\mathrm{Spin}(10)$ in $\mathrm{SU}(16)$ can only induce an inner automorphism of $\mathrm{Spin}(10)$. Meanwhile, the centralizer of $\mathrm{Spin}(10)$ in $\mathrm{U}(16)$ is $S^1$, the multiples of the identity. Thus, the normalizer of $\mathrm{Spin}(10)$ in $\mathrm{U}(16)$ is $S^1{\cdot}\mathrm{Spin}(10)$, which is connected.