In a situation like this, one usually uses symmetry to focus on the upper half of the cone, with equation $z=\sqrt{x^2+y^2}$. The area element in this case is simply $$\sqrt{1+|\nabla z|^2}\,dx\,dy = \sqrt{1+\frac{x^2+y^2}{x^2+y^2}}\,dx\,dy = \sqrt{2}\,dx\,dy \tag1$$
Integration happens over the region $x^2+y^2\le 2y$, which is a closed disk of radius $2y$.
Just kidding, it's a disk of radius $1$ with center at $(1,0)$. So the integral is $\sqrt{2}$ times $\pi 1^2$. And yes, this was a half of original surface, so the final answer is $2\sqrt{2}\pi$.
Can the area of surface formed by intersection be calculated through surface integrals?(means is there any general method to write their parametric equation)
That's the only [calculus] way to calculate surface area:
$$A(S)=\iint_A 1\,dS = \iint |r_u\times r_v|\,du\,dv$$
The above is a special case, with parameters $u$ and $v$ being $x,y$.
Is surface area of inside equal to area outside?(means area inside a ball and outside it)
I've no idea what inside and outside mean here. But the area above $z=0$ is equal to the area below it, by symmetry.
If yes then can 1. be solved by calculating area of two surfaces separately and then adding them?
That's what I did above.
The intersection of a cylinder with a plane is an ellipse. Find the semiaxes of the ellipse and you get
$$S=\pi ab$$
The minor semiaxis is always the same as the radius of the cylinder, in this case $b=r=2$.
The major semiaxis can be calculated from the angle between the plane and the cylinder axis. The angle the plane makes with the $z$ axis can be extracted from the plane normal, as the dot product gives us a cosine between two vectors, $\cos\alpha=(n_x,n_y,n_z)\cdot(0,0,1)=n_z$.
The normal of your plane can be read directly from the coefficient of the equation. An equation for a plane can be written as a dot product $\vec{n}\cdot\vec{r}=\rm const$, in your case $(1,2,1)\cdot(x,y,z)=4$. Rescale the normal to unit size and you get:
$$\vec{n}=\frac{(1,2,1)}{\sqrt 6}$$
and as we demonstrated above, also by using the dot product, the $z$ component of the normal equals the cosine of the angle with the $z$ axis:
$$\cos\alpha=\frac{1}{\sqrt 6}$$
If you draw the vertical cross section (the figure on the right), you can see a right triangle that relates the radius of the cylinder with the hypotenuse (the semiaxis):
$$a=\frac{r}{\cos\alpha}=2\sqrt 6$$
leading to the solution
$$S=4\pi \sqrt 6$$
EDIT:
Best Answer
$z=1-x-y$ so set $f(x,y)=1-x-y$ $$S=\int\int_D\sqrt{f_x^2+f_y^2+1}dxdy$$
$$S=\int\int_D\sqrt{3}dxdy$$
$$S=\sqrt{3}\int\int_Ddxdy$$
$$S=\sqrt{3}(4\pi) $$
$$S=4\sqrt{3}\pi$$
Note: $D$ is the region enclosed by the circle $x^2+y^2=4$ in $x,y$ plane.