Can photons push the source which is emitting them?

Yes.

If yes, will a more intense flashlight accelerate me more?

Yes

Does the wavelength of the light matter?

No

Is this practical for space propulsion?

Probably not

Doesn't it defy the law of momentum conservation?

No

In fact that last question is the key one, because photons do carry momentum (even though they have no mass). Photons, like all particles obey the relativistic equation:

$$ E^2= p^2c^2 + m^2c^4 $$

where for a photon the mass, $m$, is zero. That means the momentum of the photon is given by:

$$ p = \frac{E}{c} = \frac{h\nu}{c} $$

where $\nu$ is the frequency of the light. Let's suppose you have a flashlight that emits light with a power $W$ and a frequency $\nu$. The number of photons per second is the total power divided by the energy of a single photon:

$$ n = \frac{W}{h\nu} $$

The momentum change per second is the numbr of photons multiplied by the momentum of a single photon:

$$ P/sec = \frac{W}{h\nu} p = \frac{W}{h\nu} \frac{h\nu}{c} = \frac{W}{c} $$

But the rate of change of momentum is just the force, so we end up with an equation for the force created by your flashlight:

$$ F = \frac{W}{c} $$

Now you can see why I've answered your questions above as I have. The force is proportional to the flashlight power, but the fequency $\nu$ cancelled out so the frequency of the light doesn't matter. Momentum is conserved because it's the momentum carried by the photons that creates the force.

As for powering spaceships, your 1W flashlight creates a force of about $3 \times 10^{-9}$ N. You'd need a staggeringingly intense light source to power a rocket.

Can photons push the source which is emitting them?

Yes.

If yes, will a more intense flashlight accelerate me more?

Yes

Does the wavelength of the light matter?

No

Is this practical for space propulsion?

Probably not

Doesn't it defy the law of momentum conservation?

No

In fact that last question is the key one, because photons do carry momentum (even though they have no mass). Photons, like all particles obey the relativistic equation:

$$ E^2= p^2c^2 + m^2c^4 $$

where for a photon the mass, $m$, is zero. That means the momentum of the photon is given by:

$$ p = \frac{E}{c} = \frac{h\nu}{c} $$

where $\nu$ is the frequency of the light. Let's suppose you have a flashlight that emits light with a power $W$ and a frequency $\nu$. The number of photons per second is the total power divided by the energy of a single photon:

$$ n = \frac{W}{h\nu} $$

The momentum change per second is the numbr of photons multiplied by the momentum of a single photon:

$$ P/sec = \frac{W}{h\nu} p = \frac{W}{h\nu} \frac{h\nu}{c} = \frac{W}{c} $$

But the rate of change of momentum is just the force, so we end up with an equation for the force created by your flashlight:

$$ F = \frac{W}{c} $$

Now you can see why I've answered your questions above as I have. The force is proportional to the flashlight power, but the frequency $\nu$ cancels out so the frequency of the light doesn't matter. Momentum is conserved because it's the momentum carried by the photons that creates the force.

As for powering spaceships, your 1W flashlight creates a force of about $3 \times 10^{-9}$ N. You'd need a staggeringingly intense light source to power a rocket.