The argument for the first question goes as follows:

Consider the Pauli-Lubanski vector $ W_{\mu} = \epsilon_{\mu\nu\rho\sigma}P^{\nu}M^{\rho\sigma}$.
Where $P^{\mu}$ are the momenta and $M^{\mu\nu}$ are the Lorentz generators.
(The norm of this vector is a Poincare group casimir but this fact will not be needed for the argument.)

By symmetry considerations We have $W_{\mu} P^{\mu} = 0$. Now, in the case of a massless particle, a vector orthogonal to a light-like vector must be proportional to it (easy exercise). Thus
$ W^{\mu} = h P^{\mu}$, ($ h = const.$). Now, the zero component of the Pauli-Lubanski vector is given by:

$ W_{0} = \epsilon_{0\nu\rho\sigma}P^{\mu}M^{\mu\nu} = \epsilon_{abc}P^{a}M^{bc} = \mathbf{P}.\mathbf{J}$, (where the summation after the second equality is on the spatial indices only, and $\mathbf{J}$ are the rotation generators ).

Therefore the proportionality constant
$ h = \frac{W^{0}}{P^{0}}= \frac{\mathbf{P}.\mathbf{J}}{|\mathbf{P}|}$
is the helicity.

Now, on the quantum level, if we rotate by an angle of $2 \pi$ around the momentum axis, the wave function acquires a phase of:
$exp(2 \pi i\frac{\mathbf{P}}{|\mathbf{P}|}.\mathbf{J}) = exp(2 \pi i h)$.
This factor should be $\pm 1$ according to the particle statistics thus $h$ must be half integer.

As for the second question, a very powerful method to construct the gluon amplitudes is by the twistor approach.
Please see the following article by N.P. Nair for a clear exposition.

Update:

This update refers to the questions asked by user6818 in the comments:

For simplicity I'll consider the case of a photon and not gluons.

The strategy of the solution is based on the explicit construction of the angular momentum and spin of a free photon field (which depend on the polarization vectors) and showing that the above relations are satisfied for the photon field.
The photon momentum and the angular momentum densities can be obtained via the Noether theorem from the photon Lagrangian. Alternatively, it is well known that the photon linear momentum is given by the Poynting vector (proportional to) $\vec{E}\times\vec{B}$,
and it is not difficult to convince onself that the total angular momentum density is (proportional to) $\vec{x}\times \vec{E}\times\vec{B}$.

Now, the total angular momentum can be decomposed into angular and spin angular momenta (please see K.T. Hecht: quantum mechanics (page 584 equation 16))

$\vec{J} = \int d^3x (\vec{x}\times \vec{E}\times\vec{B}) =\int d^3x (\vec{E}\times\vec{A} + \sum_{i=1}^3 E_j \vec{x} \times \vec{\nabla} A_j )$

The first term on the right hand side can be interpreted as the spin and the second as the orbital angular momentum as it is proportional to the position.

Now, Neither the spin nor the orbital angular momentum densities are gauge invariant (only their sum is). But, one can argue that the total orbital angular momentum is zero because the position averages to zero, thus the total spin:

$ \vec{S} =\int d^3x (\vec{E}\times\vec{A})$

is gauge invariant:

Now, we can obseve that in canonical quantization: $[A_j, E_k] = i \delta_{jk}$, we get $[S_j, S_k] = 2i \epsilon_{jkl} S_l$. Which are the angular momentum commutation relations apart from the factor 2.

Now, by substituting the plane wave solution:

$A_k = \sum_{k,m=1,2} a_{km} \vec{\epsilon_m}(k) exp(i(\vec{k}.\vec{x}-|k|t)) +h.c.$

(The condition $\vec{\epsilon_m}(k).\vec{k} = 0$, is just a consequence of the vanishing of the sources).

We obtain:

$\vec{S} = \sum_{k,m=1,2}(-1){m} a^\dagger_{km}a_{km} \hat{k} = \sum_{k}(n_1-n_2)\hat{k}$

(where $n_1$, $n_2$ are the numbers of right and left circularly polarized photons). Thus for a single free photon, the total spin, thus the total angular momentum are aligned along or opposite to the momentum, which is the same result stated in the first part of the answer.

Secondly, the photon total spin operators exist and transform (up to a factor of two) as spin 1/2 angular momentum operators.

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