A linear graph of y against 1/x² is drawn for the function x²(y−a) = b. Find the values of a and b given that the graph passes through the points (1; 4) and (3; 8).

The given equation is $x^2(y-a) = b$.
We are told that a linear graph of $y$ against $\frac{1}{x^2}$ is drawn. This means we need to express $y$ in terms of $\frac{1}{x^2}$ in the form $Y = mX + c$, where $Y=y$ and $X=\frac{1}{x^2}$.

$$x^2(y-a) = b$$

Divide both sides by $x^2$:

$$y-a = \frac{b}{x^2}$$

Add $a$ to both sides:

$$y = \frac{b}{x^2} + a$$

This can be written as:

$$y = b\left(\frac{1}{x^2}\right) + a$$

Comparing this to $Y = mX + c$, we have:
$Y = y$
$X = \frac{1}{x^2}$
The gradient $m = b$
The y-intercept $c = a$

The graph passes through the points $(1; 4)$ and $(3; 8)$.
For the first point $(x, y) = (1, 4)$:
$X_1 = \frac{1}{x^2} = \frac{1}{1^2} = 1$
$Y_1 = y = 4$
So, the first point in $(X, Y)$ coordinates is $(1, 4)$.

For the second point $(x, y) = (3, 8)$:
$X_2 = \frac{1}{x^2} = \frac{1}{3^2} = \frac{1}{9}$
$Y_2 = y = 8$
So, the second point in $(X, Y)$ coordinates is $\left(\frac{1}{9}, 8\right)$.

The gradient $m$ is given by the formula $m = \frac{Y_2 - Y_1}{X_2 - X_1}$.

$$m = \frac{8 - 4}{\frac{1}{9} - 1}$$ $$m = \frac{4}{\frac{1}{9} - \frac{9}{9}}$$ $$m = \frac{4}{-\frac{8}{9}}$$ $$m = 4 \times \left(-\frac{9}{8}\right)$$ $$m = -\frac{36}{8}$$ $$m = -\frac{9}{2}$$

Since $m = b$, we have $b = -\frac{9}{2}$.

Using the linear equation $Y = mX + c$ and the point $(X_1, Y_1) = (1, 4)$:

$$4 = \left(-\frac{9}{2}\right)(1) + c$$ $$4 = -\frac{9}{2} + c$$ $$c = 4 + \frac{9}{2}$$ $$c = \frac{8}{2} + \frac{9}{2}$$ $$c = \frac{17}{2}$$

Since $c = a$, we have $a = \frac{17}{2}$.

The values are:
$a = \boxed{\frac{17}{2}}$
$b = \boxed{-\frac{9}{2}}$