Coming to question 1, that is, is your calculation correct?
Quoting from Wikipedia,
In solid sodium chloride, each ion is surrounded by six ions of the opposite charge as expected on electrostatic grounds. The surrounding ions are located at the vertices of a regular octahedron. In the language of close-packing, the larger chloride ions (167 pm in size) are arranged in a cubic array whereas the smaller sodium ions (116 pm) fill all the cubic gaps (octahedral voids) between them. This same basic structure is found in many other compounds and is commonly known as the $\ce{NaCl}$ structure or rock salt crystal structure. It can be represented as a face-centered cubic (fcc) lattice with a two-atom basis or as two interpenetrating face centered cubic lattices. The first atom is located at each lattice point, and the second atom is located halfway between lattice points along the fcc unit cell edge.
Also refer this LibreTexts article.
What does this mean?
For starters, $\ce{NaCl}$ is an FCC structure, with $\ce{Na+}$ occupying the octahedral voids, and $\ce{Cl-}$ occupying the FCC lattice.
Edge length in FCC unit cell = $2\sqrt{2}r$.
You get unit cell volume from here.
There are 4 octahedral voids (OV) in a unit cell, and thus 4 $\ce{Na+}$ ions.
Take them as whole spheres, and find their volumes.
$V_{\ce{Na+}}={4 \times \frac{4 \pi {r_{\ce{Na+}}}^3}{3}}$
Number of chloride ions = 8 corners + 6 face centres = $8\times\frac{1}{8}+6\times \frac{1}{2} = 4$ (we are multiplying the various positions with their individual contributions.)
$V_{\ce{Cl-}}={4 \times \frac{4 \pi {r_{\ce{Cl-}}}^3}{3}}$
Now, volume occupied by the ions = $V_{\ce{ions}}=V_{\ce{Na+}}+V_{\ce{Cl-}}=\frac{16\pi}{3}\times ({{r_{\ce{Na+}}}^3}+{r_{\ce{Cl-}}}^3)$
Volume of unit cell = $V_{\text{uc}} = (2\sqrt{2}r)^3=16\sqrt{2}r^3$,
Assuming radius of the ions is the same (that's how I was taught, it's a simplification for the calculation),
The correct result is:
$$\text{Packing fraction}=\frac{V_{\ce{ions}}}{V_{\text{uc}}}=74\text{%}$$
The $74\text{%}$ number is standard for all FCC unit cells.
So you can refer to this for the correct calculation.
Question 2, will this hold true for all alkali metal halides with an FCC structure?
Of course! It holds true for all alkali metal halides with an FCC structure.
The only exception is $\ce{CsCl}$ which has a non-closed packed structure type, but is commonly mistaken to be body-centered cubic (BCC structure).
You can read this article to know more.
Citations:
- Sodium Chloride, Wikipedia, The Free Encyclopedia, Wikimedia Foundation, 18 April 2004, en.wikipedia.org/wiki/Sodium_chloride#Solid_sodium_chloride
- R. D. Shannon (1976). "Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides". Acta Crystallogr A. 32 (5): 751–767. doi:10.1107/S0567739476001551.