We will only cover the first question of throwing a knife to a square in detail.
Choose a coordinate system so that the square is $[0,n]^2$. Let
- $p, q$ be the landing positions for knife's tip and end.
- $\theta$ be the angle between $x$-axis and knife's body. i.e.
$q - p = k_\theta \stackrel{def}{=} ( k \cos\theta, k \sin\theta )$
- For any expression, let $(\cdots)_{+} = \max(\cdots,0)$ be a short hand of its positive part.
- Let $\mathcal{E}$ be the event for the whole knife to lie within the square.
Since the square is convex, $\mathcal{E}$ is true if and only if $p , q \in [0,n]^2$.
This is equivalent to $p$ lies within a $(n - k|\cos\theta|)_{+} \times (n - k|\sin\theta|)_{+}$ rectangle. For example, when $\theta \in [0,\frac{\pi}{2}]$,
$$\begin{align}
p, q \in [0,n]^2
& \iff p \in [0,n]^2 \cap [ -k \cos\theta, n - k\cos\theta] \times [ -k \sin\theta, n - k\sin\theta]\\
& \iff p \in [0,n-k\cos\theta]\times[0,n-k\sin\theta]\end{align}$$
For general $\theta$, the conditional probability for event $\mathcal{E}$ is
$${\bf P}[\mathcal{E} | \theta ] = \left(1 - \frac{k}{n}|\cos\theta|\right)_{+}\left(1 - \frac{k}{n}|\sin\theta|\right)_{+}$$
Let $\alpha = \begin{cases}
0, & k \le n\\
\cos^{-1}\frac{n}{k}, & n < k < \sqrt{2} n\\
\frac{\pi}{4}, & \sqrt{2} n \le k
\end{cases}$, the desired probability equals to
$$\begin{align}
{\bf P}[\mathcal{E}] = &
\frac{1}{2\pi} \int_0^{2\pi} {\bf P}[\mathcal{E} | \theta ] d\theta\\
= & \frac{1}{2\pi}\int_0^{2\pi}\left(1 - \frac{k}{n}|\cos\theta|\right)_{+}\left(1 - \frac{k}{n}|\sin\theta|\right)_{+} d\theta\\
= & \frac{2}{\pi}\int_0^{\frac{\pi}{2}} \left(1 - \frac{k}{n}\cos\theta\right)_{+}\left(1 - \frac{k}{n}\sin\theta\right)_{+} d\theta\\
= & \frac{2}{\pi}\int_\alpha^{\frac{\pi}{2}-\alpha}
\left(1 - \frac{k}{n}\cos\theta\right)\left(1 - \frac{k}{n}\sin\theta\right) d\theta\\
= & \frac{2}{\pi}\left[
\theta - \frac{k}{n}(\sin\theta - \cos\theta) + \frac{k^2}{2n^2}\sin^2\theta
\right]_{\alpha}^{\frac{\pi}{2}-\alpha}\\
= & \frac{2}{\pi}\left(\frac{\pi}{2} - 2\alpha - \frac{2k}{n}(\cos\alpha - \sin\alpha) + \frac{k^2}{2n^2}\cos(2\alpha)\right)
\end{align}$$
When $k \ge \sqrt{2}n$, this evaluates to $0$ as expected. For the more
interesting case $k \le n$, this simplifies to
$${\bf P}[\mathcal{E}] = 1 - \frac{4k}{\pi n} + \frac{k^2}{\pi n^2}$$
For the remaining case $n < k < \sqrt{2}n$, we have the horrible mess
$${\bf P}[\mathcal{E}] = 1 - \frac{4}{\pi}\cos^{-1}\frac{n}{k}
- \frac{4}{\pi n}\left(n - \sqrt{k^2-n^2}\right)
+ \frac{2n^2 - k^2}{\pi n^2}
$$
Update
For the second question, choose a coordinate system so that the "circle" is $B(0,\frac{n}{2})$, the disk centered at origin with diameter $n$.
In order for $p, q \in B(0,\frac{n}{2})$, we need $p \in B(0,\frac{n}{2}) \cap B(-k_\theta,\frac{n}{2})$.
For any fixed $\theta$, the conditional probability becomes
$${\bf P}[\mathcal{E}|\theta] =
\frac{\verb/area/(B(0,\frac{n}{2}) \cap B(-k_\theta,\frac{n}{2}))}{\verb/area/(B(0,\frac{n}{2}))}
= \frac{\verb/area/(B(0,\frac{n}{2}) \cap B(-k_0,\frac{n}{2}))}{
\verb/area/(B(0,\frac{n}{2}))}
$$
Since the rightmost expression is independent of $\theta$, this is also
the probability we seek. The end result is
$${\bf P}[\mathcal{E}] =
\begin{cases}
\frac{2}{\pi}\left(\cos^{-1}\frac{k}{n} - \frac{k}{n}\sqrt{1 - \frac{k^2}{n^2}}\right), & k < n\\
0, & k \ge n
\end{cases}$$
