A box contains 90 discs numbered 1 to 90. Find the probability that a disc drawn at random bears a two-digit number or a perfect square number.

The numbers from 1 to 90 include one-digit numbers (1 to 9) and two-digit numbers (10 to 90).
Number of one-digit numbers $= 9$.
Number of two-digit numbers $= \text{Total numbers} - \text{Number of one-digit numbers}$
Number of two-digit numbers $= 90 - 9 = 81$.

The probability of drawing a two-digit number is $\boxed{\frac{9}{10}}$.

Perfect squares are numbers that can be expressed as $n^2$.
The perfect squares are:
$1^2 = 1$
$2^2 = 4$
$3^2 = 9$
$4^2 = 16$
$5^2 = 25$
$6^2 = 36$
$7^2 = 49$
$8^2 = 64$
$9^2 = 81$
($10^2 = 100$, which is greater than 90, so it's not included).
Number of perfect square numbers $= 9$.

The probability of drawing a perfect square number is $\boxed{\frac{1}{10}}$.

These are 5, 10, 15, ..., 90.
To find the count, divide the last number by 5:
Number of numbers divisible by 5 $= \frac{90}{5} = 18$.

The probability of drawing a number divisible by 5 is $\boxed{\frac{1}{5}}$.