I have the following two problems I am working on:

For this one I am using the limit comparison test, I did

f(x)= 1/(x^3-1)^1/2 / g(x) = 1/x

which ended up at x/(x^3-1)^1/2 which I can determine as x->infty is going to be greater than 0 but less than infinity

Nop. The asymptotic behaviour of the denominator here is like that of $\displaystyle x^{3\slash 2}$, so the whole thing behaves like $\displaystyle x^{-1\slash 2}\xrightarrow [x\to\infty] {} 0$ ...But you can try the same test with $\displaystyle \frac{1}{x^{3\slash 2}}$ , getting something that converges to 1. BTW, your integral converges.
(thus both either converge or diverge). I then integrated my g(x) which I chose as 1/x and got as a limit infty, thus the functions diverge. Is this correct?

The second one is:

For this one I took a similiar approach, except g(x) = x^2/sqroot(x^8) and determined f(x)/g(x) = 1

You mean the limit, of course.
, thus greater than zero but less than infty. So I integrated my g(x) down to -1/x, and as x->infty the limit is zero. No positive finite limit, thus it diverges.

Nop, again: the limit OF THE IMPROPER INTEGRAL exists and it's finite and thus THAT is the value of the improper integral. Again, this integral converges. Tonio
Hope I'm doing these right. Any help would be great thanks in advance.