Let

$$L= \lim_{x\to0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}-x}{x^3}$$

Using binomial expansion,

$$\sqrt{1+t} = 1+\frac t2-\frac{t^2}{8}+\frac{t^3}{16}+o(t^3)$$

and

$$\sqrt{1-t} = 1-\frac t2-\frac{t^2}{8}-\frac{t^3}{16}+o(t^3)$$

Subtracting,

$$\sqrt{1+t}-\sqrt{1-t} = t+\frac{t^3}{8}+o(t^3)$$

Put

$$t=\sin x$$

Hence,

$$\sqrt{1+\sin x}-\sqrt{1-\sin x} = \sin x+\frac{\sin^3x}{8}+o(x^3)$$

Using

$$\sin x=x-\frac{x^3}{6}+o(x^3),$$

we get

$$\sin^3x=x^3+o(x^3)$$

Therefore,

$$\sqrt{1+\sin x}-\sqrt{1-\sin x}-x$$

$$= \left(x-\frac{x^3}{6}\right) +\frac{x^3}{8} -x+o(x^3)$$

$$= x^3\left(-\frac16+\frac18\right)+o(x^3)$$

$$= -\frac{x^3}{24}+o(x^3)$$

Hence,

$$L=-\frac1{24}$$

Therefore,

$$48\lambda = 48\left(-\frac1{24}\right)$$

$$=-2$$

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 :

$$-1$$