Set up the subtraction in base three.

We subtract column by column from right to left.

$2 - 1 = 1$.

We need to subtract 2 from 1. Since $1 < 2$, we borrow from the next place value (the 2 in the $3^2$ place).
When we borrow 1 from the $3^2$ place, it becomes 3 in the $3^1$ place.
So, the 1 in the $3^1$ place becomes $1 + 3 = 4$.
Now subtract: $4 - 2 = 2$.

The 2 became 1 after borrowing.
$1 - 1 = 0$.

So, $212_{\text{three}} - 121_{\text{three}} = 21_{\text{three}}$.

The answer is $\boxed{21_{\text{three}}}$.

21. (b) Find the base five numeral expanded to give $(3 \times 5 \times 5) + (4 \times 1)$.

The expansion is $(3 \times 5 \times 5) + (4 \times 1)$.
This can be written as $(3 \times 5^2) + (4 \times 5^0)$.

$(3 \times 25) + (4 \times 1)$
$= 75 + 4$
$= 79_{\text{ten}}$

To convert $79_{\text{ten}}$ to base five, we repeatedly divide by 5 and record the remainders.

Reading the remainders from bottom to top, we get $304_{\text{five}}$.

The base five numeral is $\boxed{304_{\text{five}}}$.

22. (a) Simplify: $\frac{2}{5} + \frac{1}{6} \div \frac{1}{3}$

To divide by a fraction, multiply by its reciprocal.

Find a common denominator for 5 and 2, which is 10.
Convert each fraction to have the denominator 10:

The simplified expression is $\boxed{\frac{9}{10}}$.