How do I know if the product of two operators is hermitian? How can I know if the operator $\hat{B} = \hat{x} \hat{p}$ is hermitian or not? Should I check whether or not $\hat{x} \hat{p}$ commute? 
 A: I think i might give it a try:
First, we look at what 'hermitian' means for a QM operator. The operators $\hat{x}$ and $\hat{p}$ satisfy
$$\langle\psi| \hat{x} \phi\rangle = \langle \hat{x} \psi| \phi\rangle \text{ and }\langle\psi| \hat{p} \phi\rangle = \langle\hat{p} \psi| \phi\rangle \tag{1}$$  
To be hermitian, the composite operator $\hat{B}=\hat{x}\hat{p}$ has to satisfy the same hermiticity condition
$$\langle\psi| \hat{x}\hat{p} \phi\rangle = \langle \hat{x}\hat{p} \psi | \phi\rangle \tag{2}$$
So if we acknowledge that $\hat{x}\hat{p}=[\hat{x},\hat{p}]+\hat{p}\hat{x}$ and $\hat{B}^\dagger = \hat{p}^\dagger \hat{x}^\dagger$ hold, we can infer that
$$\langle\psi|\hat{B} \phi\rangle = \langle\hat{B}^\dagger \psi|\phi\rangle = \langle (\, [\hat{x},\hat{p}]+\hat{p}\hat{x}\, )^\dagger \psi |\phi\rangle = \langle[\hat{x},\hat{p}]^\dagger \psi|\phi\rangle + \langle\hat{x}^\dagger \hat{p}^\dagger \psi|\phi\rangle \tag{3}$$
Making use of the conditions in (1), we simplify $\langle \hat{x}^\dagger \hat{p}^\dagger \psi|\phi\rangle$ to $\langle\hat{x} \hat{p} \psi|\phi\rangle$, therefore we arrive at
$$\langle\psi|\hat{x}\hat{p} \phi\rangle = \langle[\hat{x},\hat{p}]^\dagger \psi|\phi\rangle + \langle\hat{x} \hat{p} \psi|\phi\rangle .$$
So your guess to check commutativity was indeed justified, as $[\hat{x}, \hat{p}]$ has to vanish for $\hat{B}$ to be Hermitian.
