You picked the "trivial" solutions - zero everywhere is, indeed, a solution.
But you only need the different sections to agree at one point.
Recap:
##\psi_I(0)=\psi_{II}(0)##
You need ##\psi_{II}(x)=A\cos(kx)+B\sin(kx)## to be 0 at x=0, which must mean that A=0, because ##\cos(0)=1##... so ##\psi_{II}(x)=B\sin(kx)## ... so far so good.
At the other end it is more complicated:
##A\sin(kL)=Ce^{-KL}##
##Ak\cos(kL)=-CKe^{-KL}##
... which appears to give you four variables in only two equations doesn't it?
But I think you'll find that k and K have to be related, so that's really only three variables.
In the end - the entire wavefunction has to be normalized ... so $$\int_0^L \psi_{II}^\star(x)\psi_{II}(x)dx + \int_L^\infty \psi_{III}^\star(x)\psi_{III}(x)dx = 1$$ which should give you the third equation.
You'll find that only specific values of k (hence K) will satisfy these conditions... (particularly for E<0) so providing discrete energy levels.
Hint: what are k and K both functions of?
Like I said before - the method of solving these last two is very similar to that for a finite square well ... which you can look up.
You should also be able to sketch the basic shape of the (amplitudes) first few bound-states (if they exist) ... eg. the first one starts at 0 when x=0, has a peak between 0 and L, is still >0 at x=L then decays exponentially from there.