Yes.

First choose $a$. You can take any $a$ such that $K = k(\sqrt{a})$ is a splitting field of $B$, so by Grunwald--Wang (or elementary congruences and sign conditions) you can enforce $\sigma(a)>0$ at all places where $B$ splits.

Now pick any $b$ such that $(\frac{a,b}{k})\cong B$. Multiplying $b$ by any norm from $K$ does not change the isomorphism class of resulting algebra. So all you need is to find an element $c\in K^\times$ whose norm down to $k$ has prescribed signs at the real places that split $B$. At those places, $K/k$ is split, so it the problem reduces to prescribing the signs of all $\sigma(c)$ for every real embedding $\sigma$ of $K$ above the real places of $k$ that split $B$. This amounts to finding a element of $K$ that lies in some open cone in $K\otimes_\mathbb{Q}\mathbb{R}$, which is always possible.

Yes.

First choose $a$. You can take any $a$ such that $K = k(\sqrt{a})$ is a splitting field of $B$, so by Grunwald--Wang (or elementary congruences and sign conditions) you can enforce $\sigma(a)>0$ at all places where $B$ splits.

Now pick any $b$ such that $(\frac{a,b}{k})\cong B$. Multiplying $b$ by any norm from $K$ does not change the isomorphism class of resulting algebra. So all you need is to find an element $c\in K^\times$ whose norm down to $k$ has prescribed signs at the real places that split $B$. At those places, $K/k$ is split, so it reduces the problem to prescribing the signs of $\sigma(c)$ for every real embedding $\sigma$ of $K$ above the real places of $k$ that split $B$. This amounts to finding an element of $K$ that lies in some open cone in $K\otimes_\mathbb{Q}\mathbb{R}$, which is always possible.