To show this is a fibration, do I just show that given $p: x \to y$ in $\text{ob}(E)$, and a map $b: F(p) = y \to y'$ then naturally the map $\Gamma: p \to p'$ has image $F(\Gamma): y \to y'$ so that $F(\Gamma) = b$.
(Here I'm using the definition in https://groupoids.org.uk/pdffiles/topgrpds-e.pdf page 262).
What is a fiber here? $F^{-1}(b)$ for a map $b$? Moreover, I'm not really clear what the second question about the exact sequence is asking.
Recalling Brown's definition of fibration:
Let $f:E\to B$ be the map of the "groupoid of $B$ under $x\in B$" where $f(p:x\to y)=y$. I believe the morphisms in $E$ between $p_1:x\to y_1$ and $p_2:x\to y_2$ are simply morphisms $g:y_1\to y_2$ in $B$ such that $g\circ p_1=p_2$, so the "evident way" for $f$ to act on morphisms is $f(g:p_1\to p_2)=g$. (Apologies for the notation. I hope it is clear enough.)
Now to show it is a fibration. Let $p:x\to y$ be an arbitrary element of $E$, and let $b:f(p)\to z$ be an arbitrary element of $B$ (since $f(p)=y$, this is $b:y\to z$). We want to find a morphism $e\in E$ from $p$ such that $f(e)=b$. Let $q:x\to z$ be the composition $b\circ p$. Then sort of tautologically, letting $e$ is the morphism from $p$ to $q$ using $b$ satisfies $f(e)=b$. Since $p,b$ were arbitrary, $E\to B$ is a fibration. (So the answer to your question is yes.)
The fiber $F_x$ over $x$ is the subgroupoid of objects that map to $x$ (the "over" is not the same "over" as in the definition of $E$) along with the morphisms that map to $1_x$. In particular, it is the set of all maps $x\to x$ as elements, but only with $1_x$ as the morphism for each $x\to x$. (If $B$ were the fundamental groupoid, the fiber would be the fundamental group at $x$.)
The exact sequence for the fibration is 7.2.9: $$F_x(p:x\to x)\to E(p:x\to x)\to B(x)\to \pi_0 F_x\to \pi_0 E\to \pi_0B$$
The groupoid $F_x$ is totally disconnected, so $\pi_0F_x$ is the set of maps $x\to x$. The map $B(x)\to\pi_0 F_x$, if you work through the definition, is sending a morphism $p:x\to x$ in $B$ to the object $p$ in $F_x$. This is surjective, so we expect $E(p:x\to x)\to B(x)$ to be trivial, and indeed the only morphism at $p$ in $E(p:x\to x)$ is the identity. Furthermore, $F_x(p)$ has only $1_x$, as mentioned before, so it is trivial, too.