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

**(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 \[ \boxed{x\leqslant -5\quad\text{or}\quad x\geqslant 1}. \]