A particle of mass m and negative charge of magnitude q travels...

Let the boundary of the magnetic field be the x-axis and the normal to it be the y-axis. The particle enters the magnetic field at the origin $(0,0)$.

The velocity of the particle makes an angle $\theta$ with the vertical (y-axis). From the figure, the velocity vector is directed downwards and to the right. Thus, the initial velocity is:
$\vec{v} = v\sin\theta \hat{i} - v\cos\theta \hat{j}$

The magnetic field is directed into the page:
$\vec{B} = -B\hat{k}$

The magnetic force on the negatively charged particle is:
$\vec{F} = -q(\vec{v} \times \vec{B}) = -q[(v\sin\theta \hat{i} - v\cos\theta \hat{j}) \times (-B\hat{k})]$
$\vec{F} = -q(-vB\sin\theta (\hat{i} \times \hat{k}) + vB\cos\theta (\hat{j} \times \hat{k}))$
$\vec{F} = -q(vB\sin\theta \hat{j} + vB\cos\theta \hat{i}) = -qvB\cos\theta \hat{i} - qvB\sin\theta \hat{j}$

The force is directed downwards and to the left, which means the center of the circular path lies in the third quadrant relative to the entry point. The particle will curve to the left and trace a major arc before exiting the magnetic field.

The radius of the circular path is given by equating the magnetic force to the centripetal force:
$qvB = \dfrac{mv^2}{R} \implies R = \dfrac{mv}{qB}$

Let $C$ be the center of the circular path, $P$ be the entry point, and $Q$ be the exit point. The boundary $PQ$ is horizontal. The radius $CP$ is perpendicular to the entry velocity $\vec{v}$. Since $\vec{v}$ makes an angle $\theta$ with the vertical, $CP$ makes an angle $\theta$ with the horizontal boundary $PQ$.

By symmetry, the radius $CQ$ also makes an angle $\theta$ with the boundary $PQ$. In the isosceles triangle $CPQ$, the angle subtended by the chord $PQ$ at the center $C$ is:
$\angle PCQ = \pi - (\theta + \theta) = \pi - 2\theta$

Since the center $C$ is below the boundary and the particle travels through the magnetic field region (also below the boundary), the path traced is the major arc of the circle. The angle subtended by this major arc at the center is:
$\text{Angle} = 2\pi - (\pi - 2\theta) = \pi + 2\theta$

The time spent by the particle inside the magnetic field is:
$t = \dfrac{\text{Distance}}{\text{Speed}} = \dfrac{R \times \text{Angle}}{v} = \dfrac{\left(\dfrac{mv}{qB}\right)(\pi + 2\theta)}{v} = \dfrac{m}{qB}(\pi + 2\theta)$

Answer: $\dfrac{m v}{q B}$, $\pi + 2\theta$, $\dfrac{m}{q B}(\pi + 2\theta)$