Since I usually prefer no-math solutions for geometry problems, here's one for question 1.
First, clamp the convex thingy between two verticals. The verticals are on springs, and will tightly press against whatever is between them.
Similarly, clamp the thingy between horizontal lines.
Then, rotate the convex thingy until the distance between the verticals is as big as it gets, or in other words, the longest diameter is horizontal. (This is important, since for this particular proof, we want to have the thing touching both the verticals at points that are on the same horizontal line.)
Now, the lines form a rectangle around the thingy. If the thingy was a triangle, it will look like this:
I have added a dotted vertical that splits the rectangle in two smaller rectangles. Each of these is exactly halved into parts inside and outside the triangle, so the area of the rectangle is exactly twice that of the triangle.
If the convex thingy has a more complex shape, it's all the better for us:
In the image, I have split the thingy along the maximal diameter, which is horizontal, because that's how we aligned the thingy.
Also, we know that there will be a point, or possibly a line segment, touching each side of the rectangle, because of how we clamped the lines. I chose a point on each horizontal that touches the thingy, and connected those points to the ends of the maximal diameter. This gave us a triangle on both sides of the maximal diameter.
As seen above, such a triangle's area is half of the encircling rectangle, so the combined area of the triangles is half of the outside rectangle.
Because the thingy is convex, we know from the definition of convex (a straight line segment connecting any two points of the thingy is entirely within the thingy), that all the triangle sides, and therefore the triangles themselves, are completely inside the thingy.
The convex thingy therefore completely contains within itself a quadrangle (or a triangle, if a side coincides with the longest diameter) that is half of the area of the outside rectangle. Thus, the convex thingy takes up at least half of the rectangle, and if the thingy's area is exactly 1, the rectangle cannot possibly have an area larger than 2.
For question 2, we can copy the rectangle that is not larger than 2, split the copy along a diagonal, and assemble a triangle out of the pieces, like @Jaap has already done.