I'm unable to solve either of these, although I tried everything I knew. I'm more interest in the answer of d). If I try to solve for r, the radius, I end up with a quadratic expression which I don't see how it helps. Also, the integrand does not seem to get simpler. Am I missing some key detail?
I'm must be missing some key theoretical point, which I'm unable to identify. Maybe another type of substitution?
For the first one, as suggested by Ninad above, you can change the pole of the polar co-ordinate system to any arbitrary point $(x_0,y_0)$ by simply replacing $x$ with $x-x_0$ and $y$ with $y-y_0$ in your polar co-ordinate conversion equations. This is usually the way to solve double integrals over circular regions where the centre of the circle is not at (0,0).
So, for your circle centred (1,0) you get: $$x-1=r\cos{\theta}$$ $$y=r\sin{\theta}$$ $$(x-1)^2+y^2=r^2$$
Now transforming your integrand, we see that it simply maps to $r^2$.
In terms of the bounds, since we have changed the location of our pole, we can simply say that $r$ goes from 0 to $\frac{\pi}{2}$, and $\theta$ goes from 0 to $2\pi$.
We get: $$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} (r^2) r dr d\theta$$ This simplifies to: $$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} r^3 dr d\theta$$ Which you can easily solve.
To answer the second question, you should know that area is given by the double integral of $f(x,y) =1 $ over the region.
If you draw the region, you will see that it is the area of intersection of an ellipse and the graph $x=\sqrt{y}$.
I would split this region into two, down the line $x=\frac{2}{\sqrt{5}}$.
This produces two double integrals.
(1) The triangle:$$\int_{0}^{\frac{2}{\sqrt{5}}} \int_{-x}^{x} 1 dy dx$$ Note that you don't need a double integral for this, and could just use the formula for the area of a triangle.
and (2) The remaining area: $$\int_{\frac{2}{\sqrt{5}}}^{2} \int_{-\sqrt{1-\frac{x^2}{4}}}^{\sqrt{1-\frac{x^2}{4}}} 1 dy dx$$ Note that this is also equal to: $$2\int_{\frac{2}{\sqrt{5}}}^{2} \int_{0}^{\sqrt{1-\frac{x^2}{4}}} 1 dy dx$$ since the function being integrated ($f(x,y)=1$) is symmetrical.
The overall area is the sum of these two integrals, which are easy to compute.