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Let $X$ be a smooth plane curve of genus $3$ (assume a smooth plane quartic) and $D$ a divisor of degree $2$ on this curve.

Assume that $\mathcal{l}(D)>0$. It means that there exists a rational function $f$ such that $div(f)+D\geqslant 0$, hence $D$ is linearly equivalent to $P+Q$ (since it is of degree $2$). Why it implies that $\mathcal{l}(D)=1$?

Another question: assume that $D$ is an odd theta characteristic (it means $2D\simeq K_X$ where $K_X$ is a canonical divisor). From the previous question it would follow that $D$ is of equivalent to $P+Q$. Why then the line passing through $P$ and $Q$ is bitangent to $X$?

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a) As you notice, you have $l(D)\geq 1$ .
On the other hand, if you had $l(D)\geq 2$, the curve $X$ would have a $g^1_2$, and thus be hyperelliptic.
But this is impossible because hyperelliptic curves (of genus $\geq 2$) cannot be embedded into the projective plane. Hence $l(D)=1$

b) The adjunction formula $$K_X=(K_{\mathbb P^2}\otimes_{\mathcal O _{\mathbb P^2}} \mathcal O_{\mathbb P^2}(X))\mid X=(\mathcal O _{\mathbb P^2}(-3)\otimes_{\mathcal O _{\mathbb P^2}} \mathcal O_{\mathbb P^2}(4))\mid X=\mathcal O _X(1)$$ shows that the canonical divisor $K_X$ consists of intersections of lines in $\mathbb P^2$ with $X$.
So in the case $2P+2Q \simeq2D\simeq K_X$, the divisor $2P+2Q$ is the intersection divisor of some line $L\subset \mathbb P^2$ with $X$.
This implies that the line through $P$ and $Q$ is tangent to $X$ at $P$ and $Q$, as required.

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