If $y=4x-5$ is tangent to the curve $y^2=p{x}^3+q$ at $(2,3)$, then:

1. Slope of Tangent: The line $y=4x-5$ has slope $m=4$.For the curve $y^2=p{x}^3+q$, differentiate w.r.t. $x$:
$$2y\frac{dy}{dx}=3p{x}^2\Rightarrow \frac{dy}{dx}=\frac{3p{x}^2}{2y}$$
At $(2,3)$:
$$\frac{3p(2{)}^2}{2(3)}=\frac{12p}{6}=2p$$
Equating slopes: $2p=4\Rightarrow p=2$.2. Point on Curve: The point $(2,3)$ must satisfy the curve $y^2=p{x}^3+q$:
$$3^2=2(2^3)+q$$
$$9=2(8)+q\Rightarrow 9=16+q\Rightarrow q=-7$$
Thus, $p=2$ and $q=-7$.