Why do we divide by $sqrt2$ in the qubit states $lvertpmrangle=frac{1}{sqrt2}(lvert0ranglepmlvert1rangle)$?

The $frac{1}{sqrt{2}}$ is due to the normalization condition which says that sum of the squares of the amplitudes of the must be equal to one while the square of the amplitude refers to the probability of getting that particular state when the qubits are measured

The vectors $|+⟩$ and $|-⟩$ are known as the eigenvectors for the Hadamard gate. When we apply the $H$ gate on the $|0⟩$ and $|1⟩$, we get $|+⟩$ and $|-⟩$ respectively.

That is, $$H|0⟩=|+⟩$$ and $$H|1⟩=|-⟩$$

#1: the $1/sqrt{2}$ is a normalization which ensures that the ``length'' of the vector is one.

#2: The notation $|pmrangle$ is just a label for the two states defined above. Since the states $|0rangle, |1rangle$ are elements of a vector space, you can take linear combinations and therefore construct the states $|pmrangle$