(ii) Fiad $$ p $$ and 9 given that $$ 2 p+3 i q+\frac{1}{1+i}-(2+i)^{2} $$, (iii) Solve the inequatity $$ |x+2| \geqslant 3 $$.

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**(ii)** We are given the equation
$$
2p+3iq+\frac{1}{1+i}-(2+i)^2=0.
$$
Assuming $$p$$ and $$q$$ are real numbers, we begin by simplifying the complex fractions and powers.

1. Compute $$(2+i)^2$$:
   $$
   (2+i)^2 = 2^2+2\cdot2\cdot i+i^2 = 4+4i-1=3+4i.
   $$

2. Simplify $$\frac{1}{1+i}$$ by multiplying the numerator and denominator by the conjugate $$1-i$$:
   $$
   \frac{1}{1+i}=\frac{1}{1+i}\cdot\frac{1-i}{1-i}=\frac{1-i}{(1+i)(1-i)}=\frac{1-i}{1^2-i^2}=\frac{1-i}{1-(-1)}=\frac{1-i}{2}=\frac{1}{2}-\frac{i}{2}.
   $$

3. Substitute these results into the original equation:
   $$
   2p+3iq+\left(\frac{1}{2}-\frac{i}{2}\right)-(3+4i)=0.
   $$
   Group the real and imaginary parts:
   
   - **Real part:**
     $$
     2p+\frac{1}{2}-3 =2p-\frac{5}{2}.
     $$
   
   - **Imaginary part:**
     $$
     3iq-\frac{i}{2}-4i=i\left(3q-\frac{1}{2}-4\right)=i\left(3q-\frac{9}{2}\right).
     $$

4. For the complex equation to be zero, both the real part and the imaginary part must vanish:
   
   - **Real part:**
     $$
     2p-\frac{5}{2}=0 \quad\Longrightarrow\quad 2p=\frac{5}{2}\quad\Longrightarrow\quad p=\frac{5}{4}.
     $$
   
   - **Imaginary part:**
     $$
     3q-\frac{9}{2}=0 \quad\Longrightarrow\quad 3q=\frac{9}{2}\quad\Longrightarrow\quad q=\frac{9}{6}=\frac{3}{2}.
     $$

Thus, 
$$
\boxed{p=\frac{5}{4},\quad q=\frac{3}{2}}.
$$

---

**(iii)** We need to solve the inequality
$$
|x+2|\geqslant 3.
$$

The absolute value inequality $$ |x+2|\geqslant 3 $$ means that the expression inside the absolute value is either at least 3 or at most $$-3$$. This gives us two cases:

1. $$ x+2\geqslant 3 $$
   $$
   x+2\geqslant 3\quad\Longrightarrow\quad x\geqslant 1.
   $$

2. $$ x+2\leqslant -3 $$
   $$
   x+2\leqslant -3\quad\Longrightarrow\quad x\leqslant -5.
   $$

Hence, the solution is 
$$
\boxed{x\leqslant -5\quad\text{or}\quad x\geqslant 1}.
$$
**(ii)** The values are $$ p = \frac{5}{4} $$ and $$ q = \frac{3}{2} $$. **(iii)** The solution to the inequality $$ |x+2| \geqslant 3 $$ is $$ x \leqslant -5 $$ or $$ x \geqslant 1 $$.

(ii) Fiad \( p \) and 9 given that \( 2 p+3 i q+\frac{1}{1+i}-(2+i)^{2} \), (iii) Solve the inequatity \( |x+2| \geqslant 3 \).

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