You know the equation of such part of the sphere is $$z^2=4-(x^2+y^2),x\in[0..2],y\in[0..2]$$ But $r^2=x^2+y^2$ and then $z=\sqrt{4-r^2}$. The ranges of our new variables are : $$\theta|_0^{\pi/2}, r|_0^2, z|_0^{\sqrt{4-r^2}}$$ So we have to evaluate $$\int_0^{\pi/2}\int_0^2\int_0^{\sqrt{4-r^2}}dv$$
There is a mistake in your cylindrical coordinates working and your spherical coordinate is wrong. Bounds of $r$ is from $0$ to $1$ and not from $0$ to $2$. None of the discs that you take will have radius more than $1$. Please also note the upper bound of $z = \sqrt{4-r^2} = 2 \,$ when $r = 0$ and $z = \sqrt{3} \,$ when $r = 1$.
So, $V = \displaystyle \int_{0}^{2\pi} \int_{0}^1 \int_{0}^{\sqrt{4-r^2}} r \, dz \, dr \, d\theta$
Now in spherical coordinates, for simplicity, your integral will be two parts - one which is spherical cone between $0 \leq \phi \leq \frac{\pi}{6}$ and then rest of the part of the cylinder which is between $\frac{\pi}{6} \leq \phi \leq \frac{\pi}{2}$.
$V = \displaystyle \int_{0}^{2\pi} \int_{0}^{\pi/6} \int_{0}^{2} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \, \, + \int_{0}^{2\pi} \int_{\pi/6}^{\pi/2} \int_{0}^{csc (\phi)} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$
You should check your sketch to understand why the second integral goes from $0$ to $\csc \phi$ (if you draw a line from the center (origin in this case) to any point on the cylinder, it is hypotenuse of a right angled triangle with angle $\phi$ to the $z$ axis and distance along $xy$ axis being $1$. So $\rho$ is $1/\sin \phi$.)
Now for the change of order of integral in cylindrical coordinates, please note $(c)$ is pretty straightforward. Now for (b), you need to use your sketch to think how to go about it - you may have to split it into two parts ($0 \leq r \leq 1, 1 \leq r \leq \sqrt{4-z^2}$). Think similarly for the other integral in spherical coordinates - how to represent $\phi$ in in terms of $\rho$.
Best Answer
I will only give an answer for the Cartesian method, to help you on your way.
For both Cartesian and Cylindrical Polar methods, you need to integrate between $z = 0$ to $z = 6$ the cross section $x^2 + y^2 = 1, y≥\frac{1}{2}$
In Cartesian, we clearly have $\frac{1}{2} ≤ y ≤1$ and then $-\sqrt{1-y^2}≤x≤\sqrt{1-y^2}$ from our equation of the circle. So our integral is thus: $$\int_{z=0}^6\int_{y=\frac{1}{2}}^1\int_{x=-\sqrt{1-y^2}}^{x=\sqrt{1-y^2}}\mathrm{d}x\mathrm{d}y\mathrm{d}z$$ I recommend making the substitution $y=\sin\theta$ to solve the integrand with the square root.
The Cylindrical Polar method is very similar: just remember your Jacobian.