Q2: Possible orders of a matrix with 36 elements

A matrix of order $m \times n$ has $m$ rows, and each row has $n$ elements. So it has $mn$ elements in all.

So each possible order is a way of writing $36$ as a product of two natural numbers $m$ and $n$⁠.

Each divisor $m$ of $36$ gives exactly one order $m \times n$⁠, with $n = 36 \div m$⁠. So the number of orders equals the number of divisors of $36$⁠.

A quick count of the divisors uses the prime factorisation $36 = 2^2 \times 3^2$⁠. Every divisor is $2^i \times 3^j$⁠, where $i$ and $j$ are each $0$⁠, $1$ or $2$⁠. That gives

Watch out: $4 \times 9$ and $9 \times 4$ are different orders (4 rows or 9 rows), so both are counted. $6 \times 6$ is counted once.