If y varies as x cubed, and y = 12 when x = 2, find the equation connecting y and x, and find y when x = 5.

The problem states that $y$ varies as $x^3$. This can be written as:

$$y = kx^3$$

where $k$ is the constant of proportionality.

We are given that $y=12$ when $x=2$. Substitute these values into the equation:

$$12 = k(2^3)$$ $$12 = k(8)$$

Divide by 8 to find $k$:

$$k = \frac{12}{8}$$ $$k = \frac{3}{2}$$

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

$$y = \frac{3}{2}x^3$$

Substitute $x=5$ into the law found in Step 3:

$$y = \frac{3}{2}(5^3)$$ $$y = \frac{3}{2}(125)$$ $$y = \frac{3 \times 125}{2}$$ $$y = \frac{375}{2}$$ $$y = 187.5$$

The law connecting $y$ and $x$ is $\boxed{y = \frac{3}{2}x^3}$.
When $x=5$, $y = \boxed{187.5}$.