How would you sketch a graph of y= -3/2cscx, I'm not really sure how you do the csc.

First note that 1 divided by a positive number which is less than gives a result which is greater than 1. In more succinct notation, if a < 1, with a positive, then 1 / a > 1.
The definition of the csc function is csc(x) = 1 / sin(x).
This means several things:

Since | sin(x) | never exceeds 1, | csc(x) | = 1 / | sin(x) | can never be less than 1.
Since | sin(x) | is almost always less than 1, | csc(x) | = 1 / | sin(x) | is almost always greater than 1.
If | sin(x) | is close to 0, then | csc(x) | = 1 / | sin(x) | is large, and the closer sin(x) gets to 0 the larger | csc(x) | will be.
As x gets close to 1, csc(x) gets close to 1/1 = 1.

Look at the graph in the text that shows sin(x) and csc(x):

Find the places where sin(x) is 1--the points where the graph of sin(x) touches the line y = 1. Notice that at those points, csc(x) also touches the graph.
Locate the x-axis intervals where sin(x) is positive--i.e., where the sine curve is above the x axis. Notice that csc(x) is also positive, which happens because 1 / sin(x) has the same sign as sin(x).
Similarly notice that csc(x) is negative when sin(x) is negative.
Notice that when sin(x) gets close to 0, the magnitude of csc(x)--i.e., its vertical distance from the x axis--gets very large. This is because the reciprocal of a number close to zero has a large magnitude; to quote the above, 'the closer sin(x) gets to 0 the larger | csc(x) | will be'.
You could also make a table for csc(theta) vs. theta, using theta values 0, pi/6, pi/4, pi/3, etc.. The values of csc(theta) would be undefined (for theta = 0), 2, sqrt(2), 2 / sqrt(3), 1, 2/sqrt(3), sqrt(2), 2, undef.
In decimal form these values would be (undef), 2, 1.4, 1.2, 1, 1.2, 1.4, 2, (undef).
You could construct the graph from a complete table.

I need help on #90 in hw. the question states: suppose that the motion of a spring is given by: d(t)=6e^-.8tcos(6pit)+4, then it says: where d is the distance, in inches, of a weight from the point at which the spring is attached to a ceiling, after t seconds. How far do you think the spring is from the ceiling when the spring stops bobbing?

The more time passes, the bigger t gets.
From first semester you know that as t becomes large, e^(-.8 t) approaches 0. This is because e^(-.8 t) = 1 / (e^(.8 t)), and the denominator will be an increasingly large power of e and will hence get very large, making its reciprocal very small.
Also | cos(6 pi t) | never gets bigger than 1, So when e^(-.8 t) is multiplied by 6, then by cos(6 pi t), the product has to approach 0.
That leaves only the + 4. As t gets large, d(t) approaches 0 + 4 or just 4. And that's how far the weight will be from the ceiling when it stops moving.

I'm not really sure how to do the problem, I tried drawing a pic, but it didn't seem to help

What is the difference in the amplitude and phase shift, I'm a little confused about how to know the difference.

Amplitude is how far the function deviates from its 'middle' value. That will be the number multiplied by the sine or the cosine.
I'm going to put off explaining the phase shift because to avoid confusing you I want to explain it in exactly the same notation as the text. So it's best you don't even read the next line until I see for sure what your text uses where I use k and d below, but here's a brief and succinct statement. I'll say more, including an explanation of what good the phase shift is, next time I get to look at the text.
If the argument of the function is k x - d, this could be expressed as k ( x - d / k) and the phase shift would be d / k.