You may recall the kinematic equation that relates final velocity, initial velocity, acceleration, and distance, respectively:

$\displaystyle v_{f}^{2}=v_{i}^{2}+2ad$

Well, for rotational motion (such as in this problem), there is a similar equation, except it relates final angular velocity, intial angular velocity, angular acceleration, and angular distance, respectively:

$\displaystyle \omega _{f}^{2}=\omega _{i}^{2}+2\alpha \theta$

The wheel starts at rest, so the initial angular velocity, $\displaystyle \omega _{i}$, is zero. The total number of revolutions of the wheel is given to be 5 revolutions. Each revolution is equivalent to an angular distance of $\displaystyle 2\pi$ radians. So, we can convert the total revolutions to an angular distance to get:

$\displaystyle \theta=5\:rev\cdot \frac{2\pi }{1\:rev}=10\pi$

The final angular velocity was given as $\displaystyle \omega$ in the text of the question. So, we should use the above equation to solve for the angular acceleration, $\displaystyle \alpha$.

$\displaystyle \omega ^{2}=0+2\alpha\cdot 10\pi$

$\displaystyle \alpha=\frac{\omega ^{2}}{20\pi }$

For a rotating object, or an object moving in a circular path, the relationship between angular acceleration and linear acceleration is

$\displaystyle a=\alpha r$

Linear acceleration is given by $\displaystyle a$, angular acceleration is $\displaystyle \alpha$, and the radius of the circular path is $\displaystyle r$.

For circular/centripetal motion, the linear acceleration is related to the object's linear velocity by

$\displaystyle a_{centripetal}=\frac{v^{2}}{r}$

We know the linear velocity is $\displaystyle 9.0 \:\frac{m}{s}$, and the radius is 1.5 m, so we can find the linear acceleration...

$\displaystyle a_{c}=\frac{(9.0\:\frac{m}{s})^{2}}{1.5\:m}=54\:\frac{m}{s^2}$

Now that we have the linear acceleration, we can use this in the equation at the top to find the angular acceleration...

$\displaystyle \alpha=\frac{a}{r}=\frac{54\:\frac{m}{s^2}}{1.5\Lm}=36\:\frac{rad}{s^2}$

The definition of angular velocity is $\displaystyle \omega=\frac{\Delta \Theta }{\Delta t}$. 

By identifying the given information to be $\displaystyle \Delta \Theta =2\pi$ and $\displaystyle \Delta t =3s$, we can plug this into the equation to calculate the angular velocity:

$\displaystyle \omega=\frac{2\pi}{3 s}=2.09s^-1$

The angular velocity of the second hand of a clock can be found by dividing the number of radians the second hand will travel over a known period of time. Thankfully for a clock, we know that the second hand will make one revolution, i.e. covering $\displaystyle 2\pi rad$ in one minute, or 60s. The formula for angular velocity is:

$\displaystyle \omega = \frac{\theta}{t}$

So the angular velocity, $\displaystyle \omega$ is $\displaystyle \frac{2\pi rad}{60s}$, which simplifies to our answer, $\displaystyle \frac{\pi rad}{30s}$

The angular velocity should not change based on the radius of the second hand. No matter what size the second hand, it will still cover one revolution every minute or 60s. The linear velocity will be greater and the angular momentum will also be greater for the clocktower, but its angular velocity will be the same. This can be seen by looking at the equation for angular velocity:

$\displaystyle \omega=\frac{\theta}{t}$

Angular velocity, in $\displaystyle \frac{rad}{s}$, is given by the length traveled divided by the time taken to travel the length:

$\displaystyle \omega=\frac{\theta}{t}$

We are told that the amount of time taken to make one revolution is 3min. One revolution is equal to $\displaystyle 2\pi\ rad$, and 3 minutes is equal to 180 seconds. Divide the radian value by the seconds value to get the angular velocity.

$\displaystyle \omega =\frac{2\pi}{180}$

$\displaystyle \omega=0.035\frac{rad}{s}$

For this question, the angular velocity $\displaystyle w$ can be given by the equation:

$\displaystyle w=\frac{\theta}{t}$, where $\displaystyle \theta$ is the angle made and $\displaystyle t$ is the time taken to make this angle. 

In this problem, the wheel makes one full revolution($\displaystyle \theta=2\pi$) in $\displaystyle 2$ seconds. 

Therefore:

$\displaystyle w=\frac{2\pi}{2s}=\pi \frac{rads}{s}$

Given initial angular velocity, angular acceleration, and time we can easily solve for final angular velocity with:

$\displaystyle \omega=\omega _{0}+\alpha t$

$\displaystyle \omega=5\frac{rad}{s}+2\frac{rad}{s} (4)={13\frac{rad}{s}}$

If the ferris wheel has height $\displaystyle 100m$ then it must have radius $\displaystyle 50m$.

The circumference of the ferris wheel, or the distance of one rotation, is then:

$\displaystyle 2\pi*r=C$

$\displaystyle 2\pi*50m=C$

$\displaystyle 2\pi*50m\approx 314m$

Convert the given velocity into meters per minute, or $\displaystyle \frac{m}{min}$:

$\displaystyle 7\frac{m}{s}*\frac{60 s}{1 min}=420\frac{m}{min}$

Find rotations per minute:

$\displaystyle 420\frac{m}{min}*\frac{1 rotation}{314m}=\frac{1.34 rotations}{minute}$

Convert $\displaystyle \frac{rotations}{minute}$ to $\displaystyle \frac{meters}{second}$:

$\displaystyle 1\frac{rotation}{minute}*\frac{1 minute}{60 seconds}*\frac{2*pi*10meters}{1 rotation}=\frac{1.05 meters}{second}$