Find the values of a and b given that lines L1 and L2 are perpendicular and pass through the points P(3,4) and Q(a,b), and Q(a,b) and R(2,-1) respectively. The gradient of L1 is -3/2.

The gradient of line L1 passing through P$(3, 4)$ and Q$(a, b)$ is given as $m_1 = -\frac{3}{2}$.
The formula for the gradient is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

$$-\frac{3}{2} = \frac{b - 4}{a - 3}$$

Cross-multiply to form the first equation:

$$2(b - 4) = -3(a - 3)$$ $$2b - 8 = -3a + 9$$ $$3a + 2b = 9 + 8$$ $$3a + 2b = 17 \quad (\text{Equation 1})$$

Line L2 is perpendicular to L1. If two lines are perpendicular, the product of their gradients is $-1$.

$$m_1 \cdot m_2 = -1$$ $$-\frac{3}{2} \cdot m_2 = -1$$ $$m_2 = \frac{-1}{-\frac{3}{2}} = \frac{2}{3}$$

Line L2 passes through points Q$(a, b)$ and R$(2, -1)$ with a gradient $m_2 = \frac{2}{3}$.
Using the gradient formula:

$$\frac{2}{3} = \frac{-1 - b}{2 - a}$$

Cross-multiply to form the second equation:

$$2(2 - a) = 3(-1 - b)$$ $$4 - 2a = -3 - 3b$$ $$2a - 3b = 4 + 3$$ $$2a - 3b = 7 \quad (\text{Equation 2})$$

We have two equations:

1. $3a + 2b = 17$
2. $2a - 3b = 7$

Multiply Equation 1 by 3 and Equation 2 by 2 to eliminate $b$:
$3 \times (3a + 2b = 17) \implies 9a + 6b = 51$
$2 \times (2a - 3b = 7) \implies 4a - 6b = 14$

Add the two new equations:

$$(9a + 6b) + (4a - 6b) = 51 + 14$$ $$13a = 65$$ $$a = \frac{65}{13}$$ $$a = 5$$

Substitute the value of $a$ into Equation 1:

$$3(5) + 2b = 17$$ $$15 + 2b = 17$$ $$2b = 17 - 15$$ $$2b = 2$$ $$b = 1$$

The values are $a = 5$ and $b = 1$.
The values of $a$ and $b$ are $\boxed{a=5, b=1}$.