For question 1, it comes down to probability. I have two distinguishable particles, $a$ and $b$. The probability density to find particle $a$ at $x_1$ is $$P_a (x_1)= \Psi_a(x_1) \Psi_a^*(x_1),$$ and we have a similar expression for particle $b$ at $x_2$. The probability density to find particle $a$ at $x_1$ and particle $b$ at $x_2$ is just the product of probability densities $P_a$, $P_b$. The probability density is then $$\Psi(x_1,x_2)\Psi^*(x_1,x_2)=\Psi_a(x_1) \Psi_a^*(x_1) \Psi_b(x_2) \Psi_b^*(x_2)$$ For any complex number the conjugate is just a multiplication by a phase away: $$(a+b i)^*=e^{i\alpha}(a+b i)$$ $\alpha$ depends on $a$ and $b$. From here I can then write $$\Psi(x_1,x_2)=\Psi_a(x_1) \Psi_b(x_2) e^{i\phi}$$ But the last phase is irrelevant, so you get just the product of individual wavefunctions.
For question 2, we go back to probability again. We know that we cannot distinguish particles $a$ and $b$. Then $$\Psi(x_1,x_2)=e^{i\phi}\Psi(x_2,x_1)$$ Repeating the same formula again for $x_2,x_1$ we get $$\Psi(x_2,x_1)=e^{i\phi}\Psi(x_1,x_2)$$. When we plug it into the previous formula, we have $$\Psi(x_1,x_2)=e^{i\phi}\Psi(x_2,x_1)=e^{2i\phi}\Psi(x_1,x_2)$$ This yields $e^{2i\phi}=1$ or $e^{i\phi}=\pm1$. Therefore $\Psi(x_1,x_2)=\pm\Psi(x_2,x_1)$. So the total wavefunction is either symmetric (+) or antisymmetric (-).
For the last question: we start saying that $\Psi(x_1,x_2)$ is a linear combination of $\Psi_a(x_1) \Psi_b(x_2)$ and $\Psi_a(x_2) \Psi_b(x_1)$, so we can write $$\Psi(x_2,x_1)=a\Psi_a(x_1) \Psi_b(x_2)+b\Psi_a(x_2) \Psi_b(x_1)$$ or the equivalent $$\Psi(x_2,x_1)=A[\Psi_a(x_1) \Psi_b(x_2)+e^{i\phi}\Psi_a(x_2) \Psi_b(x_1)]$$ In a similar fashion as for the previous question, we get that $e^{i\phi}$ has to be $+1$ or $-1$. The choice of the sign depends on the symmetry of the total wavefunction (if particles are bosons or fermions)
