By applying the second version of the Fundamental Theorem of Calculus find the integral: The second version of the Fundamental Theorem of Calculus states that if $F'(x)=f(x)$ then $\int_{a}^{b} f(x) dx = F(b)-F(a)$. 

I need to use this to find a) $\int_{-2}^{-1} \frac{1}{x^3} dx $ and b) $\int_{0}^{x} t \cdot cos(t^2) dt $

Now I can solve both integrals easily enough and get a) $-\frac{3}{8}$ and b) $\frac{1}{2}sin(x^2)$ as my answers. However I don't know what qualifies as using the Fundamental Theorem of Calculus. I've just integrated as I normally would and got my answers. Do I need to be particularly rigorous or what?
Computing these integrals "easily enough" means using the FTC. That's what the FTC is all about.

The "hard way" would be to rewrite them as limits of Riemann sums and computing these limits "directly". The latter can only be done in special cases, like the following: $\int_0^1 e^x\ dx$.
we have $\left(-\frac{1}{2}x^{-2}\right)'=\frac{1}{x^3}$ thus we get
$\int_{-2}^{-1}\frac{1}{x^3}dx=\frac{1}{2}(-2)^{-2}-\frac{1}{2}(-1)^{-2}=-\frac{3}{8}$
is it this what you mean?
sorry for my mistake, further we have
$\left(\frac{1}{2}\sin(x^2)\right)'=\frac{1}{2}\cos(x^2)2x=x\cos(x^2)$ and we get
$\int_{0}^{x}t\cos(t^2)dt=\frac{1}{2}\sin(x^2)$