For example, if I have two signals, $\cos(2\pi ft+\frac\pi4)+\cos(2\pi ft+\frac\pi3)$, what would be different in both transforms (Fourier and cosine) how would the spectrum changes?
And What would happened for a signal like $\cos(2\pi ft)+\cos(2\pi ft+0.99\pi)$.
These integral transforms are linear, which means the transform of a sum is identical to the sum of the individual transforms, i.e.
$\mathcal F\left\{\cos(a)+\cos(b)\right\}= \mathcal F\left\{\cos(a)\right\}+\mathcal F\left\{\cos(b)\right\}$,
and also, that when multiplied with a constant, that constant can be "inside" or "outside" the transform:
$\mathcal F\left\{\alpha g(t) \right\}=\alpha\mathcal F\left\{g(t) \right\}\, \forall \alpha \text{ const.}$
No magic happening here; since (Euler's formula! It's important!):
$$\begin{align} \mathcal F\left\{\cos(a)\right\}&=\mathcal F\left\{\frac 12\left(e^{i a}+e^{-ia}\right)\right\}\\ &=\frac12\mathcal F\left\{e^{i a}+e^{-ia}\right\}\quad\quad\quad\quad\quad||\,a=2\pi f t + \varphi\\ &=\frac12\mathcal F\left\{e^{i(2\pi ft + \varphi)}+e^{-i(2\pi ft + \varphi)}\right\}\\ &=\frac12\mathcal F\left\{e^{i 2\pi ft}\cdot e^{i\varphi}+e^{-i 2\pi ft }\cdot e^{-i\varphi}\right\}\\ &=\frac12\mathcal F\left\{e^{i 2\pi ft}\cdot e^{i\varphi}\right\}+\frac12\mathcal F\left\{e^{-i 2\pi ft }\cdot e^{-i\varphi}\right\}\\ &=\frac{e^{i\varphi}}2\mathcal F\left\{e^{i 2\pi ft}\right\}+\frac{e^{-i\varphi}}2\mathcal F\left\{e^{-i 2\pi ft }\right\}\tag{*}\\ \end{align}$$
we see that for any fixed phase $\varphi$ of a cosine, for example $\frac\pi 3$ or $\frac\pi 4$, the Fourier Transform "consists" of two dirac impulses $\delta$ with positions that only depend on the cosine's frequency $f$ and with amplitudes that depend only on the phase offset.
Now, just insert your different $\varphi$ in $(*)$ and add up those sums of Fourier Transforms to see the very basic point:
What you're modelling here, mathematically, is the interference of two waves at the same frequency. Tada! Physical significance added.
Now: good luck with the rest of your homework.
