Given

$$\lim_{x \to \infty}\frac{x^2f(x)}{x^4+3x^2+1}=2.$$

Since

$$x^4+3x^2+1 \sim x^4$$

for large $$x$$,

$$\lim_{x\to\infty}\frac{x^2f(x)}{x^4}=2$$

$$\lim_{x\to\infty}\frac{f(x)}{x^2}=2.$$

Therefore, $$f(x)$$ is a quadratic polynomial with leading coefficient $$2$$.

Let

$$f(x)=2x^2+bx+c.$$

Using

$$f(2)=5,$$

we get

$$8+2b+c=5$$

$$2b+c=-3.$$

Using

$$f(3)=10,$$

we get

$$18+3b+c=10$$

$$3b+c=-8.$$

Subtracting,

$$b=-5.$$

Substituting back,

$$c=7.$$

Hence,

$$f(x)=2x^2-5x+7.$$

Now,

$$\lim_{x\to1}\frac{f(x)-f(1)}{\sin(x-1)}.$$

Since

$$f(1)=2-5+7=4,$$

the limit becomes

$$\lim_{x\to1}\frac{f(x)-4}{\sin(x-1)}.$$

Using

$$\lim_{t\to0}\frac{\sin t}{t}=1,$$

we get

$$\lim_{x\to1}\frac{f(x)-f(1)}{\sin(x-1)} = \lim_{x\to1}\frac{f(x)-f(1)}{x-1}.$$

Therefore,

$$=f'(1).$$

Differentiating,

$$f'(x)=4x-5.$$

Hence,

$$f'(1)=4-5=-1.$$

Final Answer :

$$-1$$