Finding Filtered Output using Convolution property of Continuous Fourier Transform

I am not sure about intuition in general, but regarding the step-function FT being a sync function:

Note that the shape will remain the same, but the frequencies over which the FT of a particular step-function resides is a function of the pulse-width of the original signal. Namely, expanding a function in the time-domain actually shrinks the corresponding frequency-domain function (think slowing down voice recordings, the sound gets very low i.e. lower frequency).

That being said, as you decrease the pulse-width of a particular step-function the frequency components of that signal increase because now there is more change happening (to use a loose descriptor) in a shorter amount of time.

In contrast, if we expand the step-function in the time-domain to have a longer pulse-width then there is less change and the corresponding frequency components must be much lower.

In general, I look at a function and try and get a feel for how quickly it might be changing to get a rough idea. But as I said, I don't know of any general rule of thumb here.

First you need to understand that the crucial property of the delta function is that it picks a single value of a function when it gets integrated

$$ \int_{-\infty}^\infty f(x)\delta(x-a)\;dx = f(a) $$

Using this property to calculate the inverse fourier transform of \$ \pi \left[\delta(\omega+\omega_0) + \delta(\omega-\omega_0)\right] \$ you get $$ \frac 1{2\pi} \int_{-\infty}^\infty \pi\left[\delta(\omega-\omega_0) + \delta(\omega+\omega_0)\right] e^{j\omega t}d\omega = \frac 12 e^{j\omega_0 t} + \frac 12 e^{-j\omega_0 t} = \cos \omega_0 t $$ So in order to get a certain amplitude you have to multiply the delta function by some factor (weight), otherwise you get an amplitude of 1.

Since the amplitude of delta function is infinity by definition, the height is often used to indicate the weight.