Let's solve the problem step by step.

Question a:

We are tasked to find the smallest value of $k$ such that the composite function $gf(x)$ can be formed.

Key idea: For $gf(x)$ to exist, the range of $f(x)$ must fit into the domain of $g(x)$.

1. The domain of $f(x) = 2x - 3$ is $x \geq k$. Hence, $f(x) \geq 2k - 3$. The range of $f(x)$ is $[2k - 3, \infty)$.

2. The domain of $g(x) = x^2 - 4$ is $x \geq -4$. Thus, for $gf(x)$ to exist, $f(x) \geq -4$.

   Setting $f(x) \geq -4$: $2x - 3 \geq -4 \implies 2x \geq -1 \implies x \geq -\frac{1}{2}.$ But $x \geq k$. Therefore, $k \geq -\frac{1}{2}$.

3. Thus, the smallest $k$ such that $gf(x)$ can be formed is: $k = -\frac{1}{2}.$

Question b:

We need to solve the inequality $gf(x) > 45$.

1. First, substitute $f(x) = 2x - 3$ into $g(x)$: $gf(x) = g(2x - 3) = (2x - 3)^2 - 4.$

2. Solve $gf(x) > 45$: $(2x - 3)^2 - 4 > 45 \implies (2x - 3)^2 > 49.$

3. Solve $(2x - 3)^2 > 49$: $|2x - 3| > 7 \implies 2x - 3 > 7 \text{ or } 2x - 3 < -7.$

   Solve each case:

   • $2x - 3 > 7$: $2x > 10 \implies x > 5.$
   • $2x - 3 < -7$: $2x < -4 \implies x < -2.$

4. Combine with $x \geq k$. From part (a), $k = -\frac{1}{2}$, so $x \geq -\frac{1}{2}$. This restricts the solution to: $x > 5.$

Final Answers:

a) The smallest value of $k$ is $k = -\frac{1}{2}$.
b) The inequality $gf(x) > 45$ is satisfied for $x > 5$.

Would you like a detailed explanation of any of the steps? Here are five related questions to consider:

1. How is the range of $f(x)$ determined?
2. Why must $f(x)$ fit within the domain of $g(x)$?
3. How do we handle absolute value inequalities like $|2x - 3| > 7$?
4. Can $gf(x)$ ever be less than $45$? How would we solve that inequality?
5. What happens if $k > -\frac{1}{2}$?

Tip: Always ensure the domain compatibility between functions when forming composites!