Total number of discs = 90

(i) Let E₁ be the event of having a two digit number.

Number of discs bearing two digit number = 90 - 9 = 81

Let E₁ be the event of getting a good pen

Therefore P(getting  a two digit number) = P(E₁) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_1}{number\,of\,all\,possible\,outcomes}\) = \(\frac{81}{90}\) = \(\frac{9}{10}\)

Thus, the probability that the discs bears a two digit number is \(\frac{9}{10}\).

(ii) Let E₂ be the event of getting a perfect squaring number.

Disc bearing perfect square numbers are 1,4,9,16,25,36,49,64 and 81.

number of discs bearing a perfect square number = 9

Therefore P(getting  a perfect square number) = P(E₂) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_2}{number\,of\,all\,possible\,outcomes}\) = \(\frac{9}{90}\) = \(\frac{1}{10}\) 

Thus, the probability that the discs bears a perfect square number is \(\frac{1}{10}\)

(iii) Let E₃ be the event of getting a number divisible by 5.

Discs bearing numbers divisible by 5 are 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85 and 90.

number of discs bearing a number divisible by 5 = 18

Therefore P(getting  a number divisible by 5) = P(E₃) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_3}{number\,of\,all\,possible\,outcomes}\) = \(\frac{18}{90}\) = \(\frac{1}{5}\)

Thus, the probability that the discs bears a number divisible by 5 is \(\frac{1}{5}\).