1. Let $k$ be a given positive integer, and $P$ be a point in plane $\alpha$. In plane $\alpha$, if $n$ lines can be drawn, none of which pass through point $P$, such that any ray with $P$ as its endpoint and lying in this plane intersects at least $k$ of these $n$ lines, find the minimum value of $n$.
1. If at most $2 k$ lines are drawn, let one of these lines be $l_{0}$, and the two rays passing through point $P$ and parallel to $l_{0}$ have at most $2 k-1$ intersection points with the other lines. Therefore, there must be a ray that intersects these lines at most $k-1$ times. This is a contradiction.
Thus, $n \geqslant 2 k+1$.
Let $P$ be the starting point, and for a given angle $\theta$, let $s(\theta)$ be the set of rays with $P$ as the endpoint and intersecting the unit circle at point $(\cos \theta, \sin \theta)$. For each $\theta_{i}$ and $\varepsilon>0$, draw two almost parallel lines on each ray from $P$ to ensure that for some $\theta\left(\theta_{i} \leqslant \theta \leqslant \theta_{i}+\varepsilon\right)$, all rays from point $P$ intersect at least one of these two lines in $s(\theta)$.

Next, let $l$ be a line in the plane $\alpha$ that does not pass through point $P$. Take $2 k$ points $Q_{1}, Q_{2}, \cdots, Q_{2 k}$ sequentially on $l$, and take points $S_{1}, S_{2}, \cdots, S_{k}$ such that for $1 \leqslant j \leqslant k$, point $P$ is inside $\triangle S_{j} Q_{2 j-1} Q_{2 j}$.
Consider a ray $a$ in the plane $\alpha$ with $P$ as the endpoint.
For $1 \leqslant j \leqslant k$, since point $P$ is inside $\triangle S_{j} Q_{2 j-1} Q_{2 j}$, if $a$ does not intersect $Q_{2 j-1} Q_{2 j}$, then $a$ must intersect $S_{j} Q_{2 j-1}$ or $S_{j} Q_{2 j}$. And for the $k$ triangles, the $k$ segments $Q_{2 j-1} Q_{2 j}(1 \leqslant j \leqslant k)$ lie on the same line and have no common parts, so there is at most one $i(1 \leqslant i \leqslant k)$ such that $a$ intersects $Q_{2 i-1} Q_{2 i}$.

Thus, $a$ intersects exactly one side of each of these $k$ triangles, and these sides are all different.

Therefore, taking $l, S_{j} Q_{2 j-1}, S_{j} Q_{2 j}(1 \leqslant j \leqslant k)$, these $2 k+1$ lines satisfy the problem's requirements.
In conclusion, the minimum value of $n$ is $2 k+1$.