Confusion in basics of Laplace Transform

I have few confusions while starting Laplace Transform. So far I have studied, Fourier series and Fourier Transform. The basic difference which I found from different books is Fourier Transform is only considered the imaginary part whereas the Laplace transform considers both real and imaginary for general values.
i) I want to ask that, is it only difference we have in Laplace and Fourier Transform? Then i saw the two different equations of Laplace transform which is bilateral Laplace Transform $$ X(s) = \int _{-\infty}^{+\infty} x(t) e^{-st} dt$$ Whereas in the second equation of Laplace Transform is called unilateral Laplace Transform and it defined as: $$ X(s) = \int _{0}^{+\infty} x(t) e^{-st}dt $$ It omit the negative part and only have for $t>0$

ii) Here I want to ask that what is the reason of omitting the $t<0$ part?

And Lastly, there was an example $$ x(t)=e^{-at}u(t)$$ After applying Laplace transform it was written The transform exists only if $Re(s+a)$ is possitive.
iii) Now here is am confuse that why it took for possitive?

Answer

The unilateral Laplace transform is used for analyzing causal linear time-invariant systems, which have an impulse response $h(t)$ that is zero for $t<0$. The unilateral Laplace transform can be used to solve initial value problems, due to the correspondence

$$x'(t)\Longleftrightarrow sX(s)-x(0)$$

where $x(0)$ is a given initial value for the function $x(t)$. Note that for the bilateral Laplace transform the equivalent correspondence is simply $x'(t)\Longleftrightarrow sX(s)$.

Concerning the signal $x(t)=e^{-at}u(t)$, note that its Laplace transform is

$$X(s)=\int_0^{\infty}e^{-(a+s)t}dt$$

This integral only converges if the exponential decays, which results in the condition $\text{Re}(a+s)>0$. This condition defines the region of convergence (ROC) of the Laplace transform. $X(s)$ only exists for values of the complex variable $s$ satisfying $\text{Re}(s)>-\text{Re}(a)$.