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Yes.

Let $C$ be a hyperelliptic curve of genus $g$, and let $L$ be a general line bundle of degree $g+1$. By Riemann-Roch, $\dim|L| = 1$ and $|L|$ is base-point free, so the complete series $|L|$ gives a degree $g+1$ map to $\mathbb{P}^1$. Then the product of this map and the degree $2$ map $C\to \mathbb{P}^1$ gives a map $f:C\to \mathbb{P}^1 \times \mathbb{P}^1$, whose image is a curve of type $(2,g+1)$. But $\mathbb{P}^1\times \mathbb{P}^1$ is just a quadric in $\mathbb{P}^3$.

To see the map is an embedding, it will suffice to show that it is birational. Indeed, the image has arithmetic genus $g$ by adjunction on a quadric surface. But it also has geometric genus $g$ since $C$ is its normalization. Thus the image is smooth if it is reduced.

Finally we must see that the map is birational. The only way the map fails to be injective is if some divisor of $|L|$ contains a pair of points conjugate under the hyperelliptic involution. But in this case the assumption that $r(L)=1$ implies that $|L| = g_2^1 + p_1+\cdots+ p_{g-1}$, where $g_2^1$ is the hyperelliptic series and $p_1,\ldots,p_{g-1}$ are base points. Since $L$ was general, this isn't true, and we're done.

Here's a cute (although trivial) kind of partial converse: If $g$ is prime, then any smooth curve $C$ of genus $g$ which embeds in a smooth quadric is hyperelliptic.

Yes.

Let $C$ be a hyperelliptic curve of genus $g$, and let $L$ be a general line bundle of degree $g+1$. By Riemann-Roch, $\dim|L| = 1$ and $|L|$ is base-point free, so the complete series $|L|$ gives a degree $g+1$ map to $\mathbb{P}^1$. Then the product of this map and the degree $2$ map $C\to \mathbb{P}^1$ gives a map $f:C\to \mathbb{P}^1 \times \mathbb{P}^1$, whose image is a curve of type $(2,g+1)$. But $\mathbb{P}^1\times \mathbb{P}^1$ is just a quadric in $\mathbb{P}^3$.

To see the map is an embedding, it will suffice to show that it is birational. Indeed, the image has arithmetic genus $g$ by adjunction on a quadric surface. But it also has geometric genus $g$ since $C$ is its normalization. Thus the image is smooth if it is reduced.

Finally we must see that the map is birational. The only way the map fails to be injective is if some divisor of $|L|$ contains a pair of points conjugate under the hyperelliptic involution. But in this case the assumption that $\dim |L|=1$ implies that $|L| = g_2^1 + p_1+\cdots+ p_{g-1}$, where $g_2^1$ is the hyperelliptic series and $p_1,\ldots,p_{g-1}$ are base points. Since $L$ was general, this isn't true, and we're done.

Here's a cute (although trivial) kind of partial converse: If $g$ is prime, then any smooth curve $C$ of genus $g$ which embeds in a smooth quadric is hyperelliptic.

Yes.

Let $C$ be a hyperelliptic curve of genus $g$, and let $L$ be a general line bundle of degree $g+1$. By Riemann-Roch, $r(L) = 1$ and $|L|$ is base-point free, so the complete series $|L|$ gives a degree $g+1$ map to $\mathbb{P}^1$. Then the product of this map and the degree $2$ map $C\to \mathbb{P}^1$ gives a map $f:C\to \mathbb{P}^1 \times \mathbb{P}^1$, whose image is a curve of type $(2,g+1)$. But $\mathbb{P}^1\times \mathbb{P}^1$ is just a quadric in $\mathbb{P}^3$.

To see the map is an embedding, it will suffice to show that it is birational. Indeed, the image has arithmetic genus $g$ by adjunction on a quadric surface. But it also has geometric genus $g$ since $C$ is its normalization. Thus the image is smooth if it is reduced.

Finally we must see that the map is birational. The only way the map fails to be injective is if some divisor of $|L|$ contains a pair of points conjugate under the hyperelliptic involution. But in this case the assumption that $r(L)=1$ implies that $|L| = g_2^1 + p_1+\cdots+ p_{g-1}$, where $g_2^1$ is the hyperelliptic series and $p_1,\ldots,p_{g-1}$ are base points. Since $L$ was general, this isn't true, and we're done.

Here's a cute (although trivial) kind of partial converse: If $g$ is prime, then any smooth curve $C$ of genus $g$ which embeds in a smooth quadric is hyperelliptic.

Yes.

Let $C$ be a hyperelliptic curve of genus $g$, and let $L$ be a general line bundle of degree $g+1$. By Riemann-Roch, $\dim|L| = 1$ and $|L|$ is base-point free, so the complete series $|L|$ gives a degree $g+1$ map to $\mathbb{P}^1$. Then the product of this map and the degree $2$ map $C\to \mathbb{P}^1$ gives a map $f:C\to \mathbb{P}^1 \times \mathbb{P}^1$, whose image is a curve of type $(2,g+1)$. But $\mathbb{P}^1\times \mathbb{P}^1$ is just a quadric in $\mathbb{P}^3$.

To see the map is an embedding, it will suffice to show that it is birational. Indeed, the image has arithmetic genus $g$ by adjunction on a quadric surface. But it also has geometric genus $g$ since $C$ is its normalization. Thus the image is smooth if it is reduced.

Finally we must see that the map is birational. The only way the map fails to be injective is if some divisor of $|L|$ contains a pair of points conjugate under the hyperelliptic involution. But in this case the assumption that $r(L)=1$ implies that $|L| = g_2^1 + p_1+\cdots+ p_{g-1}$, where $g_2^1$ is the hyperelliptic series and $p_1,\ldots,p_{g-1}$ are base points. Since $L$ was general, this isn't true, and we're done.

Here's a cute (although trivial) kind of partial converse: If $g$ is prime, then any smooth curve $C$ of genus $g$ which embeds in a smooth quadric is hyperelliptic.

Let $C$ be a hyperelliptic curve of genus $g$, and let $L$ be a general line bundle of degree $g+1$. By Riemann-Roch, $r(L) = 1$ and $|L|$ is base-point free, so the complete series $|L|$ gives a degree $g+1$ map to $\mathbb{P}^1$. Then the product of this map and the degree $2$ map $C\to \mathbb{P}^1$ gives a map $f:C\to \mathbb{P}^1 \times \mathbb{P}^1$, whose image is a curve of type $(2,g+1)$. But $\mathbb{P}^1\times \mathbb{P}^1$ is just a quadric in $\mathbb{P}^3$.

To see the map is an embedding, it will suffice to show that it is birational. Indeed, the image has arithmetic genus $g$ by adjunction on a quadric surface. But it also has geometric genus $g$ since $C$ is its normalization. Thus the image is smooth if it is reduced.

Finally we must see that the map is birational. The only way the map fails to be injective is if some divisor of $|L|$ contains a pair of points conjugate under the hyperelliptic involution. But in this case the assumption that $r(L)=1$ implies that $|L| = g_2^1 + p_1+\cdots+ p_{g-1}$, where $g_2^1$ is the hyperelliptic series and $p_1,\ldots,p_{g-1}$ are base points. Since $L$ was general, this isn't true, and we're done.

Yes.

Let $C$ be a hyperelliptic curve of genus $g$, and let $L$ be a general line bundle of degree $g+1$. By Riemann-Roch, $r(L) = 1$ and $|L|$ is base-point free, so the complete series $|L|$ gives a degree $g+1$ map to $\mathbb{P}^1$. Then the product of this map and the degree $2$ map $C\to \mathbb{P}^1$ gives a map $f:C\to \mathbb{P}^1 \times \mathbb{P}^1$, whose image is a curve of type $(2,g+1)$. But $\mathbb{P}^1\times \mathbb{P}^1$ is just a quadric in $\mathbb{P}^3$.

To see the map is an embedding, it will suffice to show that it is birational. Indeed, the image has arithmetic genus $g$ by adjunction on a quadric surface. But it also has geometric genus $g$ since $C$ is its normalization. Thus the image is smooth if it is reduced.

Finally we must see that the map is birational. The only way the map fails to be injective is if some divisor of $|L|$ contains a pair of points conjugate under the hyperelliptic involution. But in this case the assumption that $r(L)=1$ implies that $|L| = g_2^1 + p_1+\cdots+ p_{g-1}$, where $g_2^1$ is the hyperelliptic series and $p_1,\ldots,p_{g-1}$ are base points. Since $L$ was general, this isn't true, and we're done.

Here's a cute (although trivial) kind of partial converse: If $g$ is prime, then any smooth curve $C$ of genus $g$ which embeds in a smooth quadric is hyperelliptic.