(1)
f(x)=x/2 ∀x=2⁻ᵏ: k∈ℕ
Let y=f(x)=x/2, so x=2y.
Let g(y)=x=2y, then g(x)=2x ∀x=2¹⁻ᵏ: k∈ℕ, and g(x)=f⁻¹(x).
Therefore f(x) is bijective because it has an inverse.
Range of f(x) is (0,½]. Range of g(x) is (0,1] which is the range of f⁻¹(x).
(2)
g:(0,1)→(0,1]
Consider (2¹⁻²ᵏ+2⁻ᵏ-1)/(2¹⁻ᵏ-1) for k∈ℕ.
When k=1, the expression is undefined.
As k→∞, the expression → 1.
For x=2⁻ᵏ, the expression becomes (2x²+x-1)/(2x-1)=(2x-1)(x+1)/(2x-1).
This expression reduces to x+1 for x≠½. When k=1, x=½.
Let g(x)=(2x²+x-1)/(2x-1) which has the range (0,1) and domain (0,½), since x=2⁻ᵏ: k∈ℕ.
If y=g(x)=x+1, then x=y-1, so if h(y)=x, h(y)=y-1, h(x)=x-1. And h(x)=g⁻¹(x), so g(x) has an inverse which makes g bijective.