Question 1

1/(2+i)²

Solve this question....

Question 2

$\large\underline{\sf{Solution-}}$

Given complex number is

$\rm \longmapsto\:\frac{1}{ {(2 + i)}^(2) }$

$\rm \:  =  \: \frac{1}{4 + {i}^(2) + 2 * 2 * i}$

$\rm \:  =  \: (1)/(4 - 1 + 4i)$

$\rm \:  =  \: (1)/(3+ 4i)$

$\rm \:  =  \: (1)/(3+ 4i) * (3 - 4i)/(3 - 4i)$

$\rm \:  =  \: \frac{3 - 4i}{ {3}^(2) - {(4i)}^(2) }$

$\rm \:  =  \: \frac{3 - 4i}{9 -16 {i}^(2) }$

We know,

$\rm \longmapsto\:\tt{ {i}^(2) \: = \: - \: 1 \: } \\$

So, using this, we get

$\rm \:  =  \: (3 - 4i)/(9 -16( - 1))$

$\rm \:  =  \: (3 - 4i)/(9 + 16)$

$\rm \:  =  \: (3 - 4i)/(25)$

$\rm \:  =  \: (3)/(25) - (4)/(25)i$

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