1. Interpret the functions carefully

We are given $f: \mathbb{R}\setminus\{3\} \to \mathbb{R}\setminus\{1\}, \qquad f(x)=\frac{x-2}{x-3}$ and $g: \mathbb{R}\to\mathbb{R}, \qquad g(x)=2x-3.$

We need to solve: $f^{-1}(x)+g^{-1}(x)=\frac{13}{2}.$

2. Find $f^{-1}(x)$

Let $y=\frac{x-2}{x-3}.$ Now solve for $x$ in terms of $y$:

$y(x-3)=x-2$ $yx-3y=x-2$ $yx-x=3y-2$ $x(y-1)=3y-2$ $x=\frac{3y-2}{y-1}.$

Hence, $f^{-1}(x)=\frac{3x-2}{x-1}, \qquad x\neq 1.$

3. Find $g^{-1}(x)$

Let $y=2x-3.$ Then $x=\frac{y+3}{2}.$ So, $g^{-1}(x)=\frac{x+3}{2}.$

4. Substitute into the equation

We must solve $\frac{3x-2}{x-1}+\frac{x+3}{2}=\frac{13}{2}, \qquad x\neq 1.$

Multiply throughout by $2$: $2\cdot \frac{3x-2}{x-1}+(x+3)=13$ $2\cdot \frac{3x-2}{x-1}=10-x.$

Now multiply by $(x-1)$: $2(3x-2)=(10-x)(x-1).$

Expand both sides: $6x-4=10x-10-x^2+x$ $6x-4=-x^2+11x-10.$

Bring all terms to one side: $x^2-5x+6=0.$

Factorize: $(x-2)(x-3)=0.$

So, $x=2 \quad \text{or} \quad x=3.$

Both are valid since only $x\neq 1$ is excluded.

5. Find the sum of all values of $x$

$2+3=5.$

6. Compare with the stored answer

Our derived answer is 5, which matches option B.