According to Magma it is 120, and it is an extension of $A_5$ by $C_2$ (A:=AutomorphismGroup(HyperellipticCurve(Polynomial(GF(5),[0,-1,0,0,0,1])))), and over $F_{25}$ or over $\bar F_5$ it is 240.
Edit: Hartshorne works over an algebraically closed field, so Exc. 2.5 on p.305 proves that over $\bar F_5$ the automorphism group has order 240. Explicitly, it is generated by
$\alpha: x\mapsto x+1, y\mapsto y$ of order 5,
$\beta: x\mapsto \frac{1}{x+1}, y\mapsto \frac{y}{(1+x)^3}$ of order 6,
$\gamma: x\mapsto 2x, y\mapsto \sqrt{2}y$ of order 8.
Actually, it is clear that the group they generate has order 240 and not less, because $\beta^3$ is not the hyperelliptic involution and $\gamma^4$ is. On the other hand, as Dan explains, you cannot get more than a double cover of $PGL(2,F_5)$, so this is the whole group. Over $F_5$ however, the automorphism group is generated by $\alpha, \beta$ and $\gamma^2$, and it has order 120.
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.
Best Answer
Felipe, I believe the answer here is d=g+3. To see that you can embed your curve in this degree is straightforward - just choose a generic line bundle of degree g+3 and it will work.
In the case of hyperelliptic curves, I don't think you can do better. The key point is that any special linear series on a hyperelliptic curve comes from taking a multiple of the pullback of O(1) on P^1 together with some base points. (You can find this fact in Arbarello-Cornalba-Griffiths-Harris.) Since these cannot give rise to embeddings (either have base points or the associated map factors through the hyperelliptic involution) we conclude the the embedding line bundle has no H^1. The Riemann-Roch gives g+3 as the lower bound for having 4 sections.