If y varies inversely as x, and y=1.5 when x=8, find x when y=6 and find y when x=1.2.

The problem states that $y$ varies inversely as $x$. This can be written as:

$$y = \frac{k}{x}$$

where $k$ is the constant of proportionality.

We are given that $y=1.5$ when $x=8$. Substitute these values into the equation:

$$1.5 = \frac{k}{8}$$

Multiply both sides by 8 to find $k$:

$$k = 1.5 \times 8$$ $$k = 12$$

Substitute the value of $k$ back into the variation equation from Step 1:

$$y = \frac{12}{x}$$

Substitute $y=6$ into the law found in Step 3:

$$6 = \frac{12}{x}$$

Multiply both sides by $x$:

$$6x = 12$$

Divide both sides by 6:

$$x = \frac{12}{6}$$ $$x = 2$$

Substitute $x=1.2$ into the law found in Step 3:

$$y = \frac{12}{1.2}$$ $$y = 10$$

When $y=6$, $x = \boxed{2}$.
When $x=1.2$, $y = \boxed{10}$.