Where can i see what is the thermal dissipation of a 5630 or 1W power led?
In a good first approximation, you can assume that 100% of the energy flowing into an LED gets converted to heat. Modern LEDs do have some 10-30% efficiency, so in reality, it's but 90%-70%, but for thermal design, the order of magnitude is often important to make sure a thermal design is sound. You'd, in the situation that you seem to be in, never built something that only has 10% of thermal headroom – simply because you can't even control your ambient temperature to that degree.







If you want to think about it (it really doesn't pay, usually): 

You can find out the actual efficiency by calculating the power that goes into the LED – simply forward voltage times current – and divide the energy of the emitted light by that.

In some cases, datasheets even tell you the luminous efficiency. But that's typically the case for the more reputable manufacturers (OSRAM, Cree and so on). You picked two manufacturers that I don't know. I presume they make cheap LEDs.

What's the energy of the emitted light? For monochromatic LEDs, that's pretty simple: it's the amount of energy per photon, times the amount of photons emitted per second.

Energy per photon is very simple; from physics 101 we know 

$$ E = \frac{hc}\lambda $$

with \$c=3\cdot 10^8 \frac{\text m}{\text s}\$ the speed of light, \$\lambda\$ the wavelength and \$h\$ Planck's constant, \$6.626 \cdot 10^{-34} \,\text{Js}\$.

However, in practical LEDs, the number of photons is seldom given – you just get "Luminous Flux" in Lumen (lm) instead – and that has a direct conversion to power, however only for specific wavelengths.

So, any reasonably reputable manufacturer will give you Lumen/Watt, luminous efficacy. If you take that, and compare it to the theoretic ideal efficacy for a 555nm source, namely 683 lm/W, you get your overall efficiency. 

Let's do this for the BL-LS5730A1S1UWC-0.5W:

It has a "typical" output of 55 lm (I believe exactly nothing from this datasheet, but let's use that number) over an angle of (what? that is not even specified?!) let's say half a sphere, so let's double that. That gives you an estimated efficency of about \$\frac16\$; that means that 83% of your power is converted to heat.

These datasheets are really good for nothing. At least two of the values opening angle (sr, °), luminous flux(lm) and luminous intensity (cd) should be given. I'd encourage you to buy from someone who gives you a complete datasheet, so that you're not surprised later on.