1. Write $f(x)$ piecewise

Given

We analyze it in intervals determined by the points $x=2$ and $x=5$.

• For $x<2$: \quad |x-5|=5-x,$$ so $$f(x)=(2-x)+(5-x)=7-2x.$$
• For $2\le x\le 5$: \quad |x-5|=5-x,$$ so $$f(x)=(x-2)+(5-x)=3.$$
• For $x>5$: \quad |x-5|=x-5,$$ so $$f(x)=(x-2)+(x-5)=2x-7.$$

Hence,

2. Check Statement-1: $f'(4)=0$

Since $4\in(2,5)$, from the piecewise form we have $f(x)=3 \quad \text{for } 2<x<5.$ A constant function has derivative zero, so $f'(x)=0 \quad \text{for } 2<x<5.$ Therefore, $f'(4)=0.$

So, Statement-1 is true.

3. Check Statement-2

Statement-2 says: $f$ is continuous in $[2,5]$, differentiable in $(2,5)$ and $f(2)=f(5)$.

• On $[2,5]$, we found $f(x)=3,$ so it is continuous there.
• On $(2,5)$, again $f(x)=3$, so it is differentiable there.
• Also, $f(2)=|2-2|+|2-5|=0+3=3,$ $f(5)=|5-2|+|5-5|=3+0=3.$ Hence, $f(2)=f(5).$

So, Statement-2 is true.

4. Does Statement-2 correctly explain Statement-1?

Statement-2 gives:

• continuity on $[2,5]$,
• differentiability on $(2,5)$,
• equality $f(2)=f(5)$.

From these facts, by Rolle's theorem, there exists at least one $c\in(2,5)$ such that $f'(c)=0.$ But Rolle's theorem only guarantees existence of some point $c$ in $(2,5)$, not specifically $x=4$.

The reason $f'(4)=0$ is actually stronger: on the whole interval $(2,5)$, $f(x)=3$ is constant, so derivative is zero everywhere there.

Thus, Statement-2 is not the correct explanation for Statement-1.

5. Evaluate options

• A: Statement-1 false, Statement-2 true → incorrect
• B: Both true, and Statement-2 is correct explanation → incorrect
• C: Both true, but Statement-2 is not the correct explanation → correct
• D: Statement-1 true, Statement-2 false → incorrect

Therefore, the correct option is $\boxed{\text{C}}.$