$\sin (t \omega)$ is not an Energy Signal, then how come its Fourier transform do exist? The following integral (perhaps fourier tranform of $\sin (t \omega)$ ) is not convergent:

$\int_{-\infty }^{\infty } e^{-i t \omega } \sin (t \omega ) \, dt$

As, $\sin (t \omega)$ is NOT an Energy Signal (but a Power Signal), then how come we get successful in finding the fourier transform of $\sin (t \omega)$ ?
You are right that such integrals are meaningless unless they are interpreted as distributions. And this is what we need to do, because - as you know - the Fourier transform of a sine function involves delta impulses. Let me try to make this a bit more intuitive:

The inverse Fourier transform of the delta function $\delta(\omega)$ (in the frequency domain) is given by

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega)e^{i\omega t}d\omega= \frac{1}{2\pi}e^{i0\cdot t}=\frac{1}{2\pi}$$

So we have the Fourier transform relation (time domain $\Longleftrightarrow$ frequency domain)

$$1\Longleftrightarrow 2\pi\delta(\omega)$$

Using the shifting property we obtain

$$e^{i\omega_0 t}\Longleftrightarrow 2\pi\delta(\omega-\omega_0)$$

And since

$$\sin(\omega_0t)=\frac{1}{2i}[e^{i\omega_0 t}-e^{-i\omega_0 t}]$$

we get for its Fourier transform

$$\sin(\omega_0t)\Longleftrightarrow \frac{\pi}{i}[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)]$$