b) A bullet was released from a hunter's gun with an initial velocity of 8 m/s. It then accelerates at 3.5 m/s2 for 15 seconds.

We use the equation of motion $v = u + at$, where $u$ is initial velocity, $a$ is acceleration, and $t$ is time.

$$v = 8 \text{ m/s} + (3.5 \text{ m/s}^2)(15 \text{ s})$$ $$v = 8 \text{ m/s} + 52.5 \text{ m/s}$$ $$v = 60.5 \text{ m/s}$$

The final velocity of the bullet is $\boxed{\text{60.5 m/s}}$.

We use the equation of motion $s = ut + \frac{1}{2}at^2$, where $s$ is distance, $u$ is initial velocity, $a$ is acceleration, and $t$ is time.

$$s = (8 \text{ m/s})(15 \text{ s}) + \frac{1}{2}(3.5 \text{ m/s}^2)(15 \text{ s})^2$$ $$s = 120 \text{ m} + \frac{1}{2}(3.5 \text{ m/s}^2)(225 \text{ s}^2)$$ $$s = 120 \text{ m} + (1.75 \text{ m/s}^2)(225 \text{ s}^2)$$ $$s = 120 \text{ m} + 393.75 \text{ m}$$ $$s = 513.75 \text{ m}$$

The distance traveled by the bullet is $\boxed{\text{513.75 m}}$.