Find the angular speed of the smaller gear? 
1.Why does the angular speed of small gear depend only on that of the larger gear?
2.Why does the length of linkage connecting the two gears not have any influence on the angular speed of smaller gear?
The first question can be answered by looking at slack and tightening of chain caused by rotation of the linkage.
Can someone please help to find the answer to the second question?
 A: It is unclear what you mean by the angular speed of the linkage.
If you're saying that the whole system is rotating around some axis parallel to the two axes of the gears, then to find the angular speed of any of the two gears in the resting lab frame, you can first determine each gear's angular speed in the rest frame of the system, then add the rotation of the system (paying attention to signs).
If by linkage you mean the chain, then first of all there is no obvious axis about which you might consider the chain's angular speed. You can pick any. Having picked such an axis, each small section of the chain will have its own angular speed, which will change over time.
The easiest way to grasp how a system of two gears and a chain works, is to notice that in a taut chain, the speed is constant throughout. So if you know the angular velocity $\omega_1$ and radius $r_1$ of the first pulley, then the chain's speed is $v=r_1 \omega_1$, and so if the second pulley has radius $r_2$, then its angular velocity will be:
$$ \omega_2 = \frac{v}{r_2} = \omega_1 \frac{r_1}{r_2} $$
As you can see from this equation, the angular velocities are inversely proportional to the radii. So if $r_1>r_2$ then $\omega_2>\omega_1$ and vice versa.
A: Note : After writing this answer (for which reason I am reluctant to delete it), I can now see that it is the same as @Surfcello's answer, which was posted earlier so it deserves the credit. 

The video link which you provide shows the connecting rod EG rotating about axis E while the inner wheel remains stationary. During this motion the outer wheel rotates about E and also about its own centre G. 
The belt does not slip against either wheel. It rotates around both wheels. But it is not obvious in the ground frame of reference how the rotation of the linkage EG affects the rotation of the outer wheel. 
Suppose in the ground frame (which is also the frame of the stationary inner wheel) the linkage EG is rotating anti-clockwise with angular speed $\Omega$. In a frame which is rotating with EG, the inner wheel is rotating clockwise with angular speed $\Omega$. The length of the belt between the two wheels remains fixed, so the points of contact on the two wheels move with the same linear speed. The clockwise angular speed $\omega'$ of the outer wheel in the rotating frame is therefore given by $$\Omega R=\omega' r$$ where $R, r$ are the two radii.
However, in the ground frame the outer wheel makes one complete revolution anti-clockwise if it remains stationary relative to the linkage EG whenever EG rotates by one full circle about E. So in the ground frame of reference the angular speed of the outer wheel is $$\omega = \omega'- \Omega = (\frac{R}{r}-1)\Omega$$ If the outer wheel is the larger $(r > R)$ then it will rotate in the opposite direction to the linkage EG. If the two wheels are the same size $(r=R)$ then the outer wheel does not rotate while the linkage EG rotates. (This happens, for example, when you rotate an elastic band or loop of string between two fingers by moving one around the other. The band or string does not slip against your fingers, neither do your fingers rotate about their own axes. The length of the band or string makes no difference.)
The length of the linkage EG does not affect the angular speed $\omega'$ of the outer wheel in the rotating frame of reference. And it does not affect the angular speed $\Omega$ of the linkage, which is what connects the rotation rates in the two frames of reference. So it has no effect on the angular speed of the outer wheel.
