Is angular velocity also defined for bodies not having circular motion? For example, is it defined for bodies moving in straight lines like angular momentum and torque? And, what would be the use of such definitions of these rotational motion related quantities in translational motion? Is angular displacement also defined for bodies moving in straight lines?
 A: In short : yes, it is defined also for singular bodies. Indeed, angular momentum about a point O for a particle is $\vec{L} = \vec{r}\times \vec{p}$. Where $\vec{r}$ is the position vector of the considered particle and $\vec{p}$ it's momentum.
For a rigid body, to obtain a number of equations sufficient to describe all its degrees of freedom, you will need $\vec{F} = \frac{d\vec{p}}{dt}$ AND $\frac{d\vec{L}}{dt} = \vec{r}\times \vec{F}$. The first equation will describe the translational motion of your solid , the second will describe the rotational part of its motion. These would yield 6 independent equations, that describe the 3 degrees of freedom for translational motion as well as the 3 degrees of freedom for rotational motion.
Now, for a point particle, 3 equations are sufficient to describe it's motion entirely. Hence, both $\vec{F} = \frac{d\vec{p}}{dt}$ and $\frac{d\vec{L}}{dt} = \vec{r}\times \vec{F}$ should provide the same equation of motion. Indeed, for a single particle, Newton's equation implies the second !
Indeed, taking Newton's equation and taking the cross product with $\vec{r}$ :
$$\vec{r}\times\vec{F} = \vec{r}\times\frac{d\vec{p}}{dt} = \frac{d(\vec{r}\times\vec{p})}{dt}-\frac{d\vec{r}}{dt}\times \vec{p}$$
But since $\frac{d\vec{r}}{dt}\propto \vec{v} \propto \vec{p}$, the term $\frac{d\vec{r}}{dt}\times \vec{p}$ vanishes and we get exactly the equation for angular momentum.
Edit : Just realized there was also a question about angular velocity. As angular momentum and all the other quantities, it is also defined for a single particle, about a point O. The angular velocity $\vec{\omega}$ is the vector that describes the instantaneous rotation motion of this particle about an axis passing through O. It is a vector parallel to the instantaneous axis of rotation, and therefore it is orthogonal to $\vec{r}$ ( in rotational motion, this is always the case ). Note that as $\vec{\omega}$ describes the instantaneous rotation about a point, it can only describe the component of $\vec{v}$ perpendicular to $\vec{r}$.
Knowing that, $\vec{\omega}$ satisfies $\vec{\omega}\times\vec{r}=\vec{v_{\perp}}$ (which is true for rotational motion, and hence in this case since we describe the rotational part of our total motion), where $\vec{v_{\perp}} = \vec{v}-\frac{\vec{v}\cdot\vec{r}}{|r|^2}\vec{r}$, the perpendicular velocity. Taking the cross product to the right with $\vec{r}$ yields $\underbrace{\omega\cdot \vec{r}}_{=0} \vec{r} - \vec{\omega} r^2 = \vec{v_{\perp}}\times \vec{r} = \vec{v}\times\vec{r}$ where $\omega\cdot \vec{r}$ is zero by orthogonality, and $\vec{v_{\perp}}\times\vec{r} = \vec{v}\times\vec{r}$ since $\vec{v_{\parallel}}$ does not contribute to the cross product. So $\vec{\omega}$ relative to a point O, will not be able in general to "encode" all the information of the motion. However, if we take the point to be at all times the instantaneous center of rotation of the particle, then we will have $\vec{v_{\perp}} = \vec{v}$.
This equation relates angular velocity and velocity, showing they are yet again equivalent for a single particle (i.e. either of them could be used to describe its motion), provided we choose wisely the point O of reference. This is also why, $\vec{\omega}$ is not, in general, used to describe the motion of particles. However, in rotational motions it is very useful as the instantaneous axis of rotation is stationary.
Edit, for completeness :
The requirement $\vec{\omega}\cdot\vec{r} = 0$ is actually not compatible in general with the requirement that $\vec{\omega}$ be in the direction of the axis of rotation. Usually, the origin $O$ is taken to be in the plane of the rotational motion, and in this case, $\vec{\omega}$ is indeed parallel to the axis of rotation, using the definition $\vec{\omega}r^2 = \vec{r}\times\vec{v}$. However, if we were to choose the origin outside the plane of rotation, then the resulting $\vec{\omega}$, defined as such, would not be parallel to the axis of rotation. This seems counterintuitive as in all rigid-body dynamics, the direction of the $\omega$ also instructs on the axis of rotation. 
This is why I think $\vec{\omega}r^2 = \vec{r}\times\vec{v}$ is not the best definition. For this reason, let's define $\vec{\omega}$ to be parallel to the axis of rotation, and require $\vec{\omega}\times\vec{r}=\vec{v}$. As shown, we obtain the equation $(\vec{\omega}\cdot \vec{r}) \vec{r} - \vec{\omega} r^2 = \vec{v}\times\vec{r}$.
 Taking the dot product with $\vec{r}$ yields the trivial equation $\vec{\omega}\cdot\vec{r}\vec{r}\cdot\vec{r}-\vec{\omega}\cdot\vec{r}r^2 =0$. 
The remaining part of the equations is $\vec{\omega_{\perp}}r^2 = -\vec{v}\times\vec{r}$. As we can see, the component of $\vec{\omega}$ parallel to $\vec{r}$ does not affect the value of $\vec{v}$. That is why, we can take it such that $\vec{\omega}$ is aligned with the axis of rotation. 
In conclusion, I think a more sensible definition of $\vec{\omega}$ is $\vec{\omega_{\perp}}r^2 = -\vec{v}\times\vec{r}$ and $\vec{\omega_{\parallel}}$ such that it is aligned with the axis of rotation. This gives the same exact physics as $\vec{\omega}r^2 = -\vec{v}\times\vec{r}$, but chooses an $\vec{\omega}$ that is consistent with the axis of rotation.
Of course, those two methods are valid, and coincide when we take our origin $O$ in the plane of the rotational motion, as we almost always do anyway !
To illustrate, take the rotational motion in cylindrical coordinates, with the axis of rotation along $\vec{e}_z$. The general position vector for the origin on the axis of rotation is $\vec{r} = (\rho cos(\phi),\rho sin(\phi), z)$, which gives $\vec{v} = (-\rho\omega sin(\phi),\rho \omega cos(\phi),0)$ where $\omega = \dot{\phi}$. Now let's take both roads with the two definition of $\omega$. 
With $\vec{\omega}r^2 = \vec{r}\times\vec{v}$, we would get $\vec{\omega} = (-\frac{\omega\rho z cos(\phi)}{\rho^2+z^2}, -\frac{\omega\rho z sin(\phi)}{\rho^2+z^2}, \frac{\omega\rho^2}{\rho^2+z^2})$, which is, as expected, not parallel to the rotation axis for $z\not= 0$.
With the other I suggested, we would have $\vec{\omega} = (0,0,\omega)$ (as we require $\vec{\omega}$ to be along $\vec{e}_z$ ).
They both yield $\vec{\omega}\times\vec{r} = \vec{v}$, and coincide for $z=0$.
A: 
Is angular displacement also defined for bodies moving in straight lines?

Yes. Conservation of angular momentum depends in part on due angular momentum due to translation. (The other part is body rotation.) To compute the angular velocity of some object due to translation, simply compute $\frac {\vec r \times \vec v}{r^2}$ where $\vec r$ is the position of the object relative to some origin, $\vec v$ is the velocity of the object, and $r$ is the magnitude of the position vector. This is singular when $\vec r = 0$. It is well defined at all other times.
