Order the following expressions: a = 6^(1)/(3), b = 2 sqrt(2), and c = 1.22.

$$a = 6^{\frac{1}{3}}$$

This means $a$ is the cube root of 6. We know that $1^3 = 1$ and $2^3 = 8$. So, $a$ must be between 1 and 2.
A calculator gives $a \approx 1.817$.

$$b = \log_2 \sqrt{2}$$

We can rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$.
Using the logarithm property $\log_x x^y = y$:

$$b = \log_2 2^{\frac{1}{2}}$$ $$b = \frac{1}{2}$$ $$b = 0.5$$

$$c = 1.2^2$$ $$c = 1.2 \times 1.2$$ $$c = 1.44$$

We have:
$a \approx 1.817$
$b = 0.5$
$c = 1.44$

Arranging these values in ascending order:
$0.5 < 1.44 < 1.817$
So, $b < c < a$.

The ordering of $a, b, c$ is $\boxed{b < c < a}$.