Lets assume that both your moon and the gas giant it orbits are spherical. That's likely to a real-life moon, but gas giants tend to exhibit noticable flattening (Saturn's polar radius is about 10% smaller than its equatorial radius, for example). I'm going to gloss over that anyway, because it makes the math easier.
You can assume that the orbit is circular and that rotational axis of the moon is aligned with its orbital axis because it makes the math below a tiny bit simpler, but I'll leave that detail to you.
Trigonometry time!
Your moon is centered at $M$ and has radius $MA$. Your gas giant is at $G$ and has radius $GD$. The line $MG$ is the current orbital radius of the moon.
There's two points of interest on your moon: $A$ and $B$. If you go further "counterclockwise" from $A$, you can no longer see the gas giant. As you walk "clockwise" from $A$ to $B$, the gas giant slowly rises above the horizon. As you pass point $B$ going "clockwise", you can see the whole disc of the gas giant in the sky. I've also marked $P$ as the pole of the moon, assuming its axis of rotation matches that of the gas giant, which seems reasonable.
The lines of interest are $AD$ and $BC$, which are respectively the outer and inner tangents.
The lengths of $MA$ and $Dx_o$ and $CX_i$ are all the same. You can now work out the angles: $\sin \theta = \frac{GD-MA}{MG}$ and $\sin \phi= \frac{GD+MA}{MG}$. These angles aren't latitudes, but are relative to the axis $MP$, so you'll need to convert them to latitudes and longitudes yourself.
Note that as you pass $B$ you can see the complete disc of the gas giant in the sky, but as you move from $B$ towards the equator of the moon, you get slightly closer and as such slightly less of the gas giant is visible, though the effect isn't very large. There are also portions of the polar regions of the gas giant that won't be visible... $D$ isn't visible from $B$ for example, because the curvature of the gas giant itself gets in the way.
Given your example figures, and assuming my math is right, $\theta$ is 3.33° and $\phi$ is 4.12°. This means that anywhere on the giant-facing side between latitude 85.88° north and 85.88° south will see a complete disc of gas giant in the sky. Anyone above latitude 86.67° north or below 86.67° south will always be able to see some portion of the gas giant in the sky, even if they're not on the giant-facing side of the moon. As the whole thing is rotationally symmetric, and assuming that the 0° meridian is on the subgiant point, everyone between longitude 93.33° west and 93.33° east will be able to see some portion of the gas giant regardless of latitude.
late addition
From a followup comment regarding visibility of rings: rings make for a different sort of problem entirely. You can basically treat them as a 2D object, and then you could reach for a simple horizon calculation: $\cos\lambda = \frac{r}{r + d}$ where $\lambda$ is the highest latitude that can see the ring, $r$ is the diameter of the moon and $d$ is the distance across the orbital plane. Given the likely distances involved, the rings will be technically visible from more-or-less the whole hemisphere facing the gas giant.
In practice, what happens is that the rings are very thin and the viewer is likely to be in the same orbital plane and as such they'll be barely visible... just a stripe across the planet, extending intro space on either side. What you might see but more clearly are the shadows of the rings cast upon the gas giant's surface... here's a nice photo of Titan in front of Saturn with the rings seen edge on, taken by Cassini:
(image credit ESA. Image used to be here but they've broken the page an only image search remembers)
The appearance of shadows depends on the inclination of the rings with respect to the gas giant's orbital plane, and the time of year. Their visibility is slightly more complex to calculate too, but you could always ask a different question about that one.
