Let f(x) and g(x) be twice differentiable functions satisfying...

Let $h(x) = f(x) - g(x)$.

Given that $h''(x) = 2$.

Integrating with respect to $x$, we get $h'(x) = 2x + C$.

The tangent to the curve $y = h(x)$ at $x = 3$ is $y = 4x - 5$. The slope of this tangent is $4$, which means $h'(3) = 4$.

Substituting $x = 3$ into the derivative, $h'(3) = 2(3) + C = 4 \Rightarrow 6 + C = 4 \Rightarrow C = -2$.

Thus, $h'(x) = 2x - 2$.

The point of tangency on the curve at $x = 3$ has the $y$-coordinate $y = 4(3) - 5 = 7$. Therefore, $h(3) = 7$.

Integrating $h'(x)$ with respect to $x$, we obtain $h(x) = x^2 - 2x + D$.

Using $h(3) = 7$, we have $3^2 - 2(3) + D = 7 \Rightarrow 9 - 6 + D = 7 \Rightarrow 3 + D = 7 \Rightarrow D = 4$.

So, $h(x) = x^2 - 2x + 4$.

We need to find $f(5) - g(5) = h(5)$.

Substituting $x = 5$, $h(5) = 5^2 - 2(5) + 4 = 25 - 10 + 4 = 19$.

Answer: $19$