A lot is known about this problem.

Firt of all, the answer to your first question is **no**. In fact, we can prove the following easy result:

**Proposition.** A curve $C$ of genus $3$ is hyperelliptic if and only if it contains a theta characteristic $L$ such that $h^0(L) \geq 2$.

*Proof.* Assume that $C$ is not hyperelliptic. Then on the canonical model of $C$, which is a plane quartic, an effective theta characteristic is cut out by a contact hyperplane, i.e. a bitangent line. Since $h^0(L) \geq 2$, we must have infinitely many contact hyperplanes, absurd because any plane quartic contains precisely $28$ bitangent lines.

Conversely, if $C$ is hyperelliptic then $|K_C|$ is composed with the unique $g_2^1$, and since $\deg K_C=4$ such a $g^1_2$ is a theta characteristic $L$ with $h^0(L)=2$. $\square$

For the general case, let me just state a single result, referring to 1 and the references given therein for a more complete treatment.

**Theorem.** Denote by $\mathscr{M}^r_g$ the sublocus of $\mathscr{M}_g$ consisting of curves having a theta characteristic $L$ such that $$h^0(L) \geq r+1 \quad \textrm{and} \quad h^0(L) \equiv r+1 \, (\textrm{mod } 2).$$
Then $\mathscr{M}_g^1$ (resp. $\mathscr{M}_g^2$) has pure codimension $1$ (resp. $3$) in $\mathscr{M}_{g}$ if $g \geq 3$ (resp. $g \geq 5$) and a generic point of any of its components is a curve which has only one theta-characteristic $L$ with $h^0(L)=2$ (resp. $h^0(L) =3$ if $g \geq 6$).

Moreover if $g \geq 3$ such a $L$ is not composed with an involution (resp. if $g \geq 6$ such a $L$ has no fixed points).

**References.**

1 Montserrat Teixidor I Bigas: Half canonical series on algebraic curves, *Trans. Amer. Math. Soc.* **302** (1987), 99-115.