First picture:
You got the first two right (good job!), so I'll just tackle the third one. If you have to compute [tex] f(x-4) [/tex], it means that you have to substitute [tex] x-4 [/tex] in place of [tex] x[/tex]. Differently from the first two points, this will generate a new function, rather than a specific value:
[tex] f(x) = \sqrt{x+4}+2 \implies f(x-4) = \sqrt{(x-4)+4}+2 = \sqrt{x}+2[/tex]
Second picture:
Given the point [tex] (x,y) [/tex] highlighted in the picture, you can deduce that the base is [tex] 2x [/tex] units long (since it spans from [tex] (-x,0) [/tex] to [tex] (x,0) [/tex]) and the height is [tex] y [/tex] units long (because it spans from [tex] (x,0) [/tex] to [tex] (x,y) [/tex]). So, the area of the rectangle is the multiplication between base and height:
[tex] A = 2xy [/tex]
But we know that [tex] y=f(x)=\sqrt{36-x^2} [/tex], so we have
[tex] A = 2x\sqrt{36-x^2} [/tex]
The domain of this function is given by the domain of the square root: we want its argument to be non, negative, so we have
[tex] 36-x^2 \geq 0 \iff x^2 \leq 36 \iff -6 \leq x \leq 6 [/tex]
But since the problem is symmetric, the answer is
[tex] 0 \leq x \leq 6 [/tex]
You can only see the answer [tex] 0 < x < 6 [/tex] because, if you choose [tex] x=0 [/tex] or [tex] x=6 [/tex], the rectangle degenerates to a segment, and your exercise doesn't like this scenario, apparently
