Bound state in a potential well? Reading from http://quantummechanics.ucsd.edu/ph130a/130_notes/node151.html
It says:

This means that the solutions separate into even parity and odd parity states. We could have guessed this from the potential.

Why A or B is zero, then we will have even and odd parity? What does this sentence mean?
And then, from the graph of the link above, it says:

Where the curves intersect (not including the asymptote), is an allowed energy. There is always one even solution for the 1D potential well. In the graph shown, there are 2 even and one odd solution. The wider and deeper the well, the more solutions.

The solution of the graph shows the energy. How do we know a wider and deeper well will have more solutions?
 A: 
Why A or B is zero ,then we will have even and odd parity? What is this sentence mean?

If you mean that why is it that $A$ or $B$ must be zero rather than $A$ and $B$ it is because although $A=B=0$ is also a valid solution to the equation you are discussing, it wouldn't have a physical meaning i.e. there would be no wave. It is important when working things out to step back and relate your mathematics back to the real problem you are dealing with.
If you mean what is even and odd parity then in this context it simply denotes the positive and negative phase components of the wave. Parity in a sense describes the 'handedness' of a function so that a parity transformation would switch the parity of a system which in this case would mean switching between a positive and negative exponential.

the solution of the graph shows the energy. How do we know wider and deeper well will have more solution?

If you look at the equation given that predicts the bound states you can see the right hand side is a function of $\cot(\sqrt{V_{0}} \cdot a)$. If you sketch a graph of y=cot(x) and compare it to y=cot(2x) or any larger function of cot you see that there are more solutions for each value of $x$. This is the same reason you would get more bound states as V0 indicates the depth of your well and a gives the width.
You can think of it more intuitively by remembering that as energy levels are quantized there can only be a finite amount in any range of energy. If you increase this energy range you allow for more possible states.
A: The parity operation is inversion through the origin, i.e. $\hat{P}\space f(x,y,z) = f(-x, -y, -z).$
States of well defined parity are eigenstates of this operator. Here we're interested in 1 dimension, so the two possibilities are:
$$ \hat{P}\space \psi(x) = \psi(-x) = \psi(x)$$ (even parity)
and
$$ \hat{P}\space \psi(x) = \psi(-x) = -\psi(x)$$ (odd parity)
It should be fairly obvious that $ \sin(x)$ has odd parity, and $\cos(x)$ has even parity.
As you can probably tell, parity has to do with mirror symmetry, and since our potential has mirror symmetry, you can bet our wavefunctions will have mirror symmetry too. 
(To be more precise, the parity operation commutes with our hamiltonian, and since the eigenstates of the hamiltonian are non-degenerate, these eigenstates must necessarily be eigenstates of the parity operator.)
As for the second part of your question, note the form of the right hand side:
$$\tan \left(\frac{2m(E+V_0)}{\hbar^2}a \right) $$
$\tan$ is a periodic function, and as you increase $a$ or $V_0$ (i.e. the height or depth of our well respectively), the frequency increases, and so you get more intersections with the square root on the right hand side (as more of the s shaped tan wiggles creep in from the left as the frequency gets larger). Intersections represent solutions of our equality and hence bound solutions of the Schrodinger equation.
Hope this helps. If anything is still unclear just ask and I'll try to clarify.
