The ratio of the sum of interior angles of two regular polygons, one with n sides and another with n+2 sides, is 3:4. Find the size of each exterior angle of the polygon with n sides.

The sum of the interior angles of a polygon with $N$ sides is given by $S_N = (N-2) \times 180^\circ$.

For the polygon with $n$ sides:

$$S_n = (n-2) \times 180^\circ$$

For the polygon with $n+2$ sides:

$$S_{n+2} = ((n+2)-2) \times 180^\circ = n \times 180^\circ$$

The problem states that the ratio of these sums is $3:4$.

$$\frac{S_n}{S_{n+2}} = \frac{3}{4}$$

Substitute the expressions for $S_n$ and $S_{n+2}$:

$$\frac{(n-2) \times 180^\circ}{n \times 180^\circ} = \frac{3}{4}$$

Cancel out $180^\circ$ from the numerator and denominator:

$$\frac{n-2}{n} = \frac{3}{4}$$

Cross-multiply:

$$4(n-2) = 3n$$

Distribute the 4:

$$4n - 8 = 3n$$

Subtract $3n$ from both sides:

$$n - 8 = 0$$

Add 8 to both sides:

$$n = 8$$

The polygon with $n$ sides has $n=8$ sides.
For a regular polygon, the size of each exterior angle is given by the formula $\frac{360^\circ}{N}$, where $N$ is the number of sides.

$$\text{Exterior angle} = \frac{360^\circ}{n}$$

Substitute $n=8$:

$$\text{Exterior angle} = \frac{360^\circ}{8}$$ $$\text{Exterior angle} = 45^\circ$$

The size of each exterior angle of the polygon with $n$ sides is $\boxed{\text{}45^\circ\text{}}$