Are our 'observations' of or near a black hole affected by gravity induced time dilation? This answer made me wonder if matter/energy/events near a black hole are viewed (by us) in a very different time frame, how might that affect our measurements?

The people on the smaller planet would observe the people on the extremely large planet 
moving in "slow motion". (note: due to gravitational time dilation)


On a related note, does light being 'bent' around a black hole experience time dilation or red/blue shift from our viewing perspective? (as possibly suggested by this answer).
The Schwarzschild black hole is the simplest, being just an isolated, nonrotated, uncharged black hole. If $\tau$ is the proper time measured by a clock moving along an arbitrary path in this spacetime, then in Schwarzschild coordinates
$$-\mathrm{d}\tau^2 = -\left(1-\frac{R}{r}\right)\mathrm{d}t^2 + \left(1-\frac{R}{r}\right)^{-1}\mathrm{d}r^2 + r^2\left(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\varphi^2\right)\text{,}$$
where $R = 2GM/c^2$ is the Schwarzschild radius. If the observer is stationary, then neither the radial nor angular coordinates change ($\mathrm{d}r = \mathrm{d}\theta = \mathrm{d}\varphi = 0$). So, for example, relative the rate of time of stationary observers at some radial coordinates $r_1$ and $r_2$ is just
$$\frac{\mathrm{d}\tau_1}{\mathrm{d}\tau_2} = \frac{\mathrm{d}\tau_1/\mathrm{d}t}{\mathrm{d}\tau_2/\mathrm{d}t} = \sqrt{\frac{1-R/r_1}{1-R/r_2}}\text{.}$$


  how might that affect our measurements?


In the sense of the measurement process, it doesn't, since a measurement apparatus is not time-dilated relative to itself. In the sense of what results we get, then of course it does.


  So it would logically follow that events happening near a black hole (a very large gravitational field) would seem to us to happen much more slowly than if they happened outside that field.


Yes. If the first stationary observer is shifted closer to the horizon ($r_1\to R$), then $(1-R/r_1)\to 0$, so the above time dilation goes to $0$ as well. So, for example, if that observer shines a laser beam toward the other, it will be redshifted when it arrives to the other.

This is really the same thing as time dilation. The laser beam has a certain frequency, so it takes a certain amount of time for one oscillation. Thus a frequency shift is just another way of talking about time dilation, except that time dilation means that this happens to every process, not just light: e.g, if the first observer taps his foot at a certain frequency (say, a tap per second), the second observer will see him tapping at a lower frequency (say, a tap per two seconds). Slowed. We can can still call this "redshift" in the sense that the frequency was shifted in the same way, although the term is usually used specifically for electromagnetic radiation.


  On a related note, does light being 'bent' around a black hole experience time dilation or red/blue shift from our viewing perspective?


Light itself experiences only trivial time dilation in the sense that the proper time along any lightlike trajectory is exactly $0$. Above, we were talking about stationary observers, and light is not stationary. Rather, $\mathrm{d}\tau = 0$ for light beams, always. So strictly speaking, light doesn't experience anything at all.

However, if you mean what the stationary observers would get when measuring light bent around the black hole, then it is blueshifted when coming closer to the black hole and redshifted when moving away--in the sense that this would be what stationary observers closer or farther away would measure. The details of the path it took between the observers do not matter; the ratio of measured frequencies will just be the inverse of the time dilation factor between the observers.