We need to find $S_1$, $S_2$, and then compute $S_1\cup S_2$.

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1. Find $S_1$

Given

$$S_1=\left\{x\in \mathbb{R}-\{1,2\}:\frac{(x+2)(x^2+3x+5)}{-2+3x-x^2}\ge 0\right\}.$$

Step 1: Simplify the denominator

$$-2+3x-x^2=-(x^2-3x+2)=-(x-1)(x-2).$$

So the inequality becomes

$$\frac{(x+2)(x^2+3x+5)}{-(x-1)(x-2)}\ge 0.$$

Equivalently,

$$\frac{(x+2)(x^2+3x+5)}{(x-1)(x-2)}\le 0.$$

Step 2: Analyze $x^2+3x+5$

Check its discriminant:

$$\Delta=3^2-4\cdot 1\cdot 5=9-20=-11<0.$$

Since leading coefficient is positive, we have

$$x^2+3x+5>0\quad \text{for all }x\in \mathbb{R}.$$

Thus the sign depends only on

$$\frac{x+2}{(x-1)(x-2)}\le 0,$$

with $x\ne 1,2$.

Step 3: Critical points

Critical points are

$$-2,\ 1,\ 2.$$

Now do sign analysis:

• For $x< -2$: take $x=-3$

  $$\frac{x+2}{(x-1)(x-2)}=\frac{-1}{(-4)(-5)}<0$$

  so included.

• For $-2<x<1$: take $x=0$

  $$\frac{2}{(-1)(-2)}>0$$

  so not included.

• For $1<x<2$: take $x=\frac32$

  $$\frac{\frac72}{(\frac12)(-\frac12)}<0$$

  so included.

• For $x>2$: take $x=3$

  $$\frac{5}{(2)(1)}>0$$

  so not included.

At $x=-2$, numerator is zero, denominator nonzero, so equality holds and $x=-2$ is included.

Therefore,

$$S_1=(-\infty,-2]\cup(1,2).$$

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2. Find $S_2$

Given

$$S_2=\left\{x\in \mathbb{R}:3^{2x}-3^{x+1}-3^{x+2}+27\le 0\right\}.$$

Step 1: Substitute $t=3^x$

Since $3^x>0$, let

$$t=3^x>0.$$

Then

$$3^{2x}=t^2, \quad 3^{x+1}=3t, \quad 3^{x+2}=9t.$$

So the inequality becomes

$$t^2-3t-9t+27\le 0$$ $$t^2-12t+27\le 0.$$

Step 2: Factorize

$$t^2-12t+27=(t-3)(t-9).$$

So

$$(t-3)(t-9)\le 0.$$

Hence

$$3\le t\le 9.$$

Now $t=3^x$, so

$$3\le 3^x\le 9=3^2.$$

Since $3^x$ is increasing,

$$1\le x\le 2.$$

Therefore,

$$S_2=[1,2].$$

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3. Compute $S_1\cup S_2$

We have

$$S_1=(-\infty,-2]\cup(1,2),$$ $$S_2=[1,2].$$

Therefore,

$$S_1\cup S_2 = (-\infty,-2]\cup[1,2].$$

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4. Match with options

This is Option B.

$$\boxed{(-\infty,-2]\cup[1,2]}$$