Let $X = \mathbb{P}^{3}$ and $\mathcal{E}$ and $\mathcal{G}$ locally free sheaves on $X$ such that $\text{rank}(\mathcal{E}) = e$ and $\text{rank}(\mathcal{G}) = g$.
Definition: The degeneracy scheme $\text{Sing}(\psi)$ of the $\psi : \mathcal{E} \longrightarrow \mathcal{G}$ is the zero scheme of the associated global section $\omega_{\psi} \in H^{0}(X, \bigwedge^{g}(\mathcal{E}^{\vee}) \otimes \text{det}(\mathcal{G}))$.
If $\psi : \mathcal{E} \longrightarrow \mathcal{G}$ is a generically surjective morphism, then:
1) Is it possible to have $\text{Sing}(\psi) = \emptyset$?
2) Is it possible to have $\text{Sing}(\psi) = \lbrace p_{1}, \dots, p_{k} \rbrace$?
Thanks in advance.
Question 1 has been resolved in the comments: take $\psi$ to be the identity morphism of $\mathcal{O}_{\Bbb P^n}$.
For 2, we use an alternate definition of the degeneracy locus which will be easier to work with. We define the degeneracy locus as the subscheme cut out locally by the maximal minors of the matrix which locally describe the morphism. In fact, we can pick a morphism which is globally determined by a matrix. As the ideal of a finite set of $M$ points in $\Bbb P^n$ can be written down using $n$ polynomials $f_1,\cdots,f_n$ of degree $M$ (see here, for instance), we let $\mathcal{E}=\mathcal{O}^n$ and $\mathcal{G}=\mathcal{O}(M)$ and describe the morphism globally by the matrix $\begin{pmatrix} f_1 & \cdots & f_n\end{pmatrix}$. This is clearly generically surjective (if any of the $f_i$ are invertible at a point, it's surjective there, and this is a generic condition). The common zero locus of the maximal minors (aka the $f_i$) is just $V(f_1,\cdots,f_n)=\{p_1,\cdots,p_M\}$.
This latter idea works in general: if $I$ is an ideal generated by several homogeneous polynomials $f_i$ of the same degree $d$ so that $V(I)$ is a proper subvariety, then we can express $V(I)$ as a degeneracy locus of the morphism $\mathcal{O}^n\to\mathcal{O}(d)$ where the map on each coordinate is just given by multiplication by $f_i$.