Find the base five numeral expanded to give (3 x 5 x 5) + (4 x 1) and workout 212three - 121three.

The given expression is $(3 \times 5 \times 5) + (4 \times 1)$.
This can be written as:

In base five, a number $d_n d_{n-1} \dots d_1 d_0$ is represented as $d_n \times 5^n + d_{n-1} \times 5^{n-1} + \dots + d_1 \times 5^1 + d_0 \times 5^0$.
Comparing this to our expression:
The coefficient for $5^2$ is 3.
There is no $5^1$ term, so its coefficient is 0.
The coefficient for $5^0$ (which is 1) is 4.

The digits are 3 (for $5^2$), 0 (for $5^1$), and 4 (for $5^0$).
Therefore, the base five numeral is $304_5$.

The base five numeral is $\boxed{304_5}$.

: Workout: $212_{\text{three}} - 121_{\text{three}}$

We subtract column by column from right to left.

$2_{\text{three}} - 1_{\text{three}} = 1_{\text{three}}$.

We need to calculate $1_{\text{three}} - 2_{\text{three}}$. Since $1 < 2$, we must borrow from the next place value (the $3^2$ place).
Borrow 1 from the $2_{\text{three}}$ in the $3^2$ place. This borrowed 1 is equivalent to $3_{\text{three}}$ in the $3^1$ place.
So, the $1_{\text{three}}$ in the $3^1$ place becomes $1_{\text{three}} + 3_{\text{three}} = 4_{\text{three}}$.
Now, subtract: $4_{\text{three}} - 2_{\text{three}} = 2_{\text{three}}$.
The $2_{\text{three}}$ in the $3^2$ place becomes $1_{\text{three}}$ after borrowing.

We now have $1_{\text{three}} - 1_{\text{three}} = 0_{\text{three}}$.

The result is $021_{\text{three}}$, which is written as $21_{\text{three}}$.

The result of the workout is $\boxed{21_{\text{three}}}$.