Since I like math, let's throw some math into this. I'll try to keep it as simple as possible though.
Kerr Black Holes
A rotating black hole is known as a Kerr Black Hole (named after Roy Kerr who found the numerical solution to GR equations for rotating black holes). In the case of a rotating black hole, there are two important parameters used to describe the black hole. The first is of course the mass of the black hole $M$. The second is the spin $a$. Really $a$ isn't the spin itself $-$ it's defined by $a=J/M$ (see footnote) where $J$ is the angular momentum of the black hole $-$ but it's a good proxy for spin so often you'll see scientists get lazy and just call it the spin of the black hole. The mathematics will tell you that Kerr black holes have the limitation that
$$0 \le a/M \le 1$$
Black Hole Event Horizon
The important parameter we want to calculate is the black hole's radius. If you run through the math, you find that this radius is given by
$$r_{\text{e}} = M+(M^2-a^2)^{1/2}$$
In the case when $a/M=0$ (and thus $a=0$), this reduces to just $r_{\text{e}}=2M$, or in regular units (instead of geometrized units) $r_{\text{e}}=2GM/c^2$. Hopefully you can see that this just reduces to the normal Schwarzschild radius for a non-rotating black hole and thus the equation above is a generalization to account for spin. Let's look at the other limit when $a/M=1$ (and thus $a=M$). In this case, you find that the radius is $r_{\text{e}}=M$. When $a/M=1$, you have a maximally rotating black hole, and your radius is half the normal Schwarzschild radius of a non-rotating black hole. This equation defines the radius of the Event Horizon, the point after which there is no returning from the black hole.
Ergosphere
As it turns out, when you define your equation to calculate the radius of the black hole, there are actually multiple solutions! The section above shows one such solution, but there's another important solution as well. This radius, sometimes called the static limit is given by the equation
$$r_{\text{s}} = M+\left(M-a^2\cos^2(\theta)\right)^{1/2}$$
Notice that this is almost exactly the same as above, except for that extra $\cos^2(\theta)$. This defines a different, slightly larger and somewhat "pumpkin-shaped" horizon which encompasses the inner Event Horizon defined above. The region between this outer horizon and the inner horizon is known as the Ergosphere. Without getting into the nitty gritty details, I'll just say that one important point about the Ergosphere is that anything inside it (that is, $r_{\text{e}}<r<r_{\text{s}}$) must rotate exactly with the black hole - it is physically impossible to stay still here!
Answers
They stopped short of saying that the tangential velocity of this spin rate is "c" (and how can a singularity have a "tangential velocity"?)
When you talk about the tangential velocity, there are multiple components of this black hole you/they may be talking about. One such tangential velocity is the tangential velocity of the event horizon (defined by $r_{\text{e}}$ above). We can take a look at the case of a maximally rotating black hole and say that the angular momentum, based on the equations above, of such a black hole is given by
$$J_{\text{max}} = a_{\text{max}}Mc = M^2c$$
Note that I've dropped the geometrized units just to be completely explicit. This has introduced an extra $c$ now. Remember that $a_{\text{max}}$ is achieved when $a/M=1$.
We can also define the angular momentum using the standard equation from physics 101, $J=rMv_\perp$, where of course $r$ is the radius of your object, and $v_\perp$ is the perpendicular, or else tangential, velocity of your spinning object. Recall from above that for a maximally rotating black hole, $r_{\text{e}}=M$ so we also have that
$$J_{\text{max}} = r_{\text{e}}Mv_{\perp} = M^2v_\perp$$
You can see that these two equations for $J_{\text{max}}$ only equal each other if the tangential velocity $v_\perp$ is equal to the speed of light $c$. So yes, you are correct to presume that at the fastest possible rotations, the black hole's event horizon is rotating at the speed of light!
I said though that there are multiple components you could talk about when discussing rotating black holes. The other, as you allude to, is the rotating singularity. You correctly point out - "how can a singularity have a tangential velocity"? As it turns out, Kerr black holes don't have point singularities, they have ring singularities. These are "rings" of mass with zero width but some finite radius. Almost like a disk of no height. These rings of course can then have a tangential velocity. You were correct to be suspicious of a point singularity having tangential velocity though. That is not possible.
They did say that the event horizon at maximum spin of a stellar black hole is about 1-1/2 km. and that if a black hole were to spin faster then the result would be a "naked black hole" that would defy the laws of physics (GR).
We know the equation exactly, since I defined it above. The radius of a stellar black hole (that is a black hole with mass exactly equal to the mass of the Sun, $M_\odot$) is given by
$$r=\frac{GM_{\odot}}{c^2}=1.48\:\text{km}$$
So yes, they were correct on their radius. They also state that spinning any faster results in a naked singularity. This is entirely true. To see this, go back to the equation for the event horizon. Remember that our upper spin limit is that $a=M$. What happens to our event horizon radius when $a>M$ (and thus $a/M>1$)? For arguments say, lets say $a=2M$. Then our event horizon radius becomes
$$r_{\text{e}}=M-(M^2-a^2)^{1/2}=M-(M^2-4M^2)^{1/2}=M-(-3M^2)^{1/2}=M-i\sqrt{3}M$$
Suddenly our radius is complex and has an imaginary component! That means it is not physical and thus cannot exist. Now that we don't have an event horizon, our singularity can't hide behind it and is "naked", exposed to the universe for anyone to see. GR tells us such an event shouldn't be allowed to happen because it results in all sorts of violations of physics. So somehow, something has to prevent black holes from spinning faster than a maximal black hole.
Shouldn't all black holes spin extremely fast (conservation of angular momentum) or would a retrograde accretion disk slow it down.
Yes, that is true in general. All black holes should spin extremely fast, simply because of conservation of angular momentum. In fact, I don't think I can come up with a case where a black hole was found to be not spinning. Shown below is a plot from this Nature paper which shows the measured spin of 19 supermassive black holes. They're all spinning pretty fast with some of them almost at the speed of light. None of them are even close to not spinning.
Footnote: In GR, to make the math easier, often scientists adopt special units known as geometrized units. These are units chosen in just such a way that the gravitational constant, $G$, and the speed of light, $c$, are equal to one. There's infinitely many units which allow this. Essentially this means no GR equations have $G$ or $c$ in them, but they're implicitly there, they're just equal to one and so not shown.
