How the Fourier transform of a cosine signal is existed?
The Fourier Transform is defined only for functions that are absolutely integrable or in other words $f\in\mathcal{L}_1$, where $\mathcal{L}_1$ is the set of all absolutely integrable functions. 

If you want to be mathematically rigorous, then you should assume that the Fourier transform of the cosine does not exist, as it is not integrable. Apart from it, the Dirac's delta does not fulfill the axioms of a function in modern mathematics, so you cannot say that a Fourier Transform of a function is something that is not a function. Instead of that, the Dirac's delta is defined as a measure that can be used in the Lebesgue integral, that satisfies certain axioms(https://en.wikipedia.org/wiki/Dirac_delta_function). However, the use of the cosine is valid for the common modulation operation, and is mathematically consistent (meaning that its Fourier transform exist):

$\int_{-\infty}^{\infty} |x(t)cos(2\pi t)e^{-i2\pi ft}|dt\leq \int_{-\infty}^{\infty} |x(t)|dt &lt; \infty $

However, if you are just interested in the engineering or physics point of view, you can use the Dirac delta "function" as a trick that works as the measure in order to get the operations you are interested to do. For the sinc function and square integrable functions, a Fourier Transform over the $\mathcal{L}_2$ space, which is the space of square integrable functions, is defined so that you can get the Fourier Transforms for those functions(http://math.mit.edu/~jerison/103/handouts/fourierint1.13.pdf), but if you are just interested in the engineering point of view, it is not important to be so rigorous mathematically.