Momentum of particle in a box

There are two different issues. One of them is the sign of the momentum; the other one is whether the momentum is spread (it's not because of the unnatural boundary conditions).

Concerning the first point, the standing wave (sine) is a real function and every real wave function has the same probability to carry momentum $+p$ and $-p$. So indeed, both of them are equally likely.

But even if you write the sine as a difference of two complex exponentials, it's still true that these exponentials aren't equal to the wave function everywhere – just inside the box – so it's still untrue that the momentum is sharply confined to two values $p$ and $-p$.

To get the probabilities of different momenta, you need to Fourier transform the standing wave – a few waves of the sine. One has $$\int_0^1 dx\,\sin(n\pi x)\exp(ipx) = \frac{n\pi[-1 +e^{ip}(-1)^n]}{p^2-n^2\pi^2} $$ Square the absolute value to get the probability density that the momentum is $p$. The momentum $p$ should have the natural prefactors $\hbar/L$ etc. and the overall wave function should get another normalization factor for the overall probability to equal one. This changes nothing about the shape of the probability distribution: almost all values of $p$ have a nonzero probability.

I think this is a great question. http://arxiv.org/abs/quant-ph/0103153 This article explains why we shouldn't enforce the boundary conditions that we do (the wavefunction goes to 0 at the boundaries) and instead should use the condition that the wavefunction is equal at both end points. The justification is partly for mathematical reasons, but partly because that condition is too strong physically; the wavefunction isn't measurable. On the other hand the probability of finding the particle between a and b is measurable. We just want to make sure that if a=0 and b approaches 0 that the probability approach 0 continuously. This is achievable even if the wavefunction is discontinuous.

Once we enforce the weaker condition, certain functions that are exponentials inside the box and zero outside are allowed (the ones with the same wavelengths as the energy eigenstates) and these are in fact the momentum eigenvalues. So exactly as you said, if you measure the momentum, the particle will collapse into one of those states.