If I'm stationary on my bicycle and spin the pedals backwards, will I not fall because of angular momentum? If I'm on my bicycle and they are stationary, and I spin the pedals backwards, will it be harder for me to fall sideways because of the angular momentum I'm generating with my legs and pedals?
I never notice any difference in stability and wonder why.
 A: When you fall down from the bicycle, you are either rotated to the left or to the right. That is, your angular displacement is in the plane perpendicular to the plane of the bicycle. The direction of torque vector is always perpendicular to the plane of angular displacement. Therefore, the direction of the torque causing you to fall is either in the direction in which the bicycle is pointing or the direction antiparallel to it, depending on whether you fall to the left or right.
When you spin the pedal backward, the angular displacement of your feet is in the plane of the bicycle. So, the torque you apply is perpendicular to the plane of the bicycle. Therefore, the torque causing you to fall and the torque you apply are perpendicular to each other. Since the perpendicular component of a vector is always 0, pedaling backward will not affect your stability at all.     
A: No it does not. I considered the dynamics of a titling spinning disk of radius $R$ and mass $m$, with tilt angle $\psi$ and rotational speed $\Omega$.
The the tilt angle equation of motion does not depend on the disk spin rate (or its acceleration)
$$\ddot{\psi} = \frac{4 g \sin \psi}{5 R}$$

How?
I set as x-axis along the bicycle and y-axis "up". The kinematics of the center of the disk is
$$ \begin{align} 
  {\bf r}_C& = \pmatrix{0 & R \cos\psi & R \sin \psi}^\top \\
  {\bf v}_C & = \pmatrix{0 & -R \dot{\psi} \sin\psi & R \dot{\psi} \cos \psi}^\top \\
  {\bf a}_C & = \pmatrix{0 & -R \dot{\psi}^2 \cos\psi-R \ddot{\psi} \sin\psi &  -R \dot{\psi}^2 \sin \psi+R \ddot{\psi} \cos \psi}^\top \\
\end{align} $$
This helps us find the reaction forces ${\bf N}$ from the ground using Newton's laws
$$ \left. \vphantom{\pmatrix{M\\M\\M}}  {\bf N} + m {\bf g} = m {\bf a}_C \right\} 
\begin{align} 
  N_x & = 0 \\
  N_y & = m g -m R ( \dot{\psi}^2 \cos\psi + \ddot{\psi} \sin \psi ) \\
  N_z & = m R (-\dot{\psi}^2 \sin \psi + \ddot{\psi} \cos\psi)
\end{align}$$
Now we realize that the mass moment of inertia of the disk along the tilt axis (x-axis) is $I_x = \frac{m}{4} R^2$. At the same time the angular motion of the disk is
$$\begin{align} 
{\boldsymbol \omega} &= \pmatrix{ \dot{\psi} & -\Omega \sin \psi & \Omega \cos \psi}^\top \\
{\boldsymbol \alpha} &= \pmatrix{ \ddot{\psi} & -\Omega \dot{\psi} \cos \psi & \Omega \dot{\psi} \sin \psi}^\top
\end{align}$$
Now we can find the tilt axis motion $\ddot{\psi}$ as well as the moment reaction from the ground ${\bf M}$ using Euler's rotational equations along the x-axis. Since the mass moment of inertia along x-axis does not depend on the tilt angle there are no gyroscopic effects along this axis (The ${\boldsymbol \omega} \times {\rm I}{\boldsymbol \omega}$ term has zero for its first component)
$$ N_y (R \sin \psi) + N_z (-R \cos \psi) = \left( \frac{m}{4} R^2 \right) \ddot{\psi} $$
Now you can solve the above for $\ddot{\psi}$.
A: I agree whith the other answers on the conclusion, there is no stability when stationary, backpedaling or not. However i think I can explain better why.
First you need to understand why a rolling bicycle is stable. You may have heard of the gyroscopic effect being responsible of the stability. That is only part of the answer.
When a bicycle is rolling forward, both wheels are turning and ready to provide a gyroscopic effect. Also the pedals and chain will be turning (though more slowly) and contribute. But contribute how?

When you lean sideways on your bike
Let's call longitudinal axis 'x', lateral axis pointing right 'y' and vertical axis 'z'.
When you lean sideways (displace your body along '+y'), gravity will pull you down along '-z', and ground reaction (along '+z') will impart a torque on your bike (around '-x'), making it lean further. If this continues, you will fall off.
Enter the gyroscopic effect. When the rapidly rotating wheels (rotation around '-y') receive an off-axis torque ('-x'), it reacts by tilting its rotation axis around the third axis ('+z'). But that doesn't help in itself because the bike is tilting around '-x' !
Impact of the rear wheel's and chain's gyroscopic effect
Those two elements cause the bike frame to want to rotate to the right ('+z'). Since both tyres are far apart and are well adhering to the ground, the bike is firmly planted, and that does not cause the bike to actually rotate at all (the tyres will negate this torque by transmitting it fully to the ground).
This is what happens when you're sitting on a stationary bike and are back-pedaling (although the gyroscopic effect is reversed from back-pedaling, the overall futility is conserved).
Impact of the front wheel's gyroscopic effect
The front wheel also tries to rotate right ('+z'). However, since the front wheel is hinged around the vertical the steering tube, such torque is not transmitted to the rest of the frame. Instead, the entire front wheel assembly is steered right. The front tyre will resist it a bit, but it has no lever of arm around 'z', and the fact that it is rolling along will help it deform appropriately. Now this steering changes stuff.
The bicycle is now naturally leaning in a turn, steering into it, therefore it makes the turn. Comes a point where the turning rate is sufficient to cause the 'centrifugal' force to appear and straighten the bike. From there comes the stabilizing effect.

Now you can clearly see why when standing still, with the front wheel still, no turning rate is achieved, no 'centrifigual force' is generated, and there is no stability. Adding gyroscopic effect on the chain by backpedaling will have no efffect on this process.
