for (3) it's trivial to find the vector normal to the plane (you can read the coordinates right off)
It's also trivial to see that the point $(0,0,4)$ lies in the plane.
So draw the normal vector such that in passes through this point and then sketch the plane that is normal to it.
(5) doesn't make sense as written. It looks like they have the $dy~dx$ reversed. You can't integrate over a variable, such as $y$, and then have that variable also appear in the limits.
Find out what the problem really is.
let's assume that they just have $dy~dx$ reversed so the integral is
$\displaystyle \int \limits_0^2 \int \limits_{\frac y 2}^{\frac{4-y}{2}}~2~dx~dy$
for the inner integral we see that $x$ ranges from $\dfrac y 2 \to \dfrac{4-y}{2}$
while $y$ ranges from $0 \to 2$
with a bit of plotting you can see that this is a horizontal isosceles triangle with vertices $(0,0), ~(0,4),~(1,2)$
To evaluate $I$ without integrating we note that it is simply twice the area of the integration region. I'll let you figure out what the area of this triangle is.
You should be able to integrate this using integration. It's a trivial integral.
Once you've established that the integration region is the triangle described above it's obvious that the integral with order of integration switched will be
$\displaystyle \int \limits_0^1 \int \limits_{2x}^{4-2x}~2~dy~dx$
You can compute them explicitly by setting
$x = \dfrac y 2 \Rightarrow y = 2x$
$x = \dfrac{4-y}{2} \Rightarrow y = 4-2x$
The upper limit of $x$ is where these two lines intersect.
$2x = 4-2x \Rightarrow x = 1$
In the first quadrant, x, y, and 0 are positive
$\displaystyle x\ge 0$
$\displaystyle y\ge 0$
$\displaystyle z\ge 0$
The plane $\displaystyle \frac{x}{2}+ \frac{y}{3}+ \frac{z}{4}= 1$ seperates "> 1" from "< 1". Since (0, 0, 0) is on the side that makes a bounded region, it must be "< 1"
$\displaystyle \frac{x}{2}+ \frac{y}{3}+ \frac{z}{4}\le 1$
We can write $\displaystyle 4x^2+ 4y^2= z^2$ as $\displaystyle 4x^2+ 4y^2- z^2= 0$. Adding that to $\displaystyle x^2+ y^2+ z^2= 20$ gives $\displaystyle 5x^2+ 5y^2= 20$, $\displaystyle x^2+ y^2= 4$ which is a circle with center at (0, 0) and radius 2. When $\displaystyle x^2+ y^2= 4$, $\displaystyle 4x^2+ 4y^2= 16= z^2$ so $\displaystyle z= \pm 4$. That is, $\displaystyle 4x^2+ 4y^2= z^2$ is a cone that cuts the sphere in the circles $\displaystyle x^2+ y^2= 4$, $\displaystyle z= \pm 4$. The given region is the portion inside the cone that is in the sphere.