probability that a square does not cover a corner of a unit square 
Suppose the Cartesian plane is tiled with an infinite tiling of unit squares. If another unit square is dropped onto the plane at random with position and orientation independent of the tiling, determine, with proof, the probability that it does not cover any corners of the squares of the tiling.

Set coordinates so that the tiling includes the (filled) square $T = \{(x,y) : 0\leq x, y \leq 1\}$. It is then equivalent to choose the second square by first choosing a point uniformly at random in $T$ to be the center of the square and then choosing an angle of rotation uniformly at random from the interval $[0,\frac{\pi}2]$. For each $\theta \in [0,\pi/2]$, circumscribe a square $T_\theta$ around $T$ with angle of rotation $\theta$ relative to $T$.

Can someone formally prove why the side lengths of $T_\theta$ are $\sin \theta + \cos\theta$? I get that rotations preserve angles so if one rotated a unit square an angle of $\theta$ about a vertex then corresponding side lengths would clearly form an angle of $\theta$, but this is slightly different.

Now inside $T_\theta$, draw the smaller square $T_\theta'$ consisting of points at distance at least $\frac{1}2$ from each side of $T_\theta$. Then $T_\theta'$ has side length $\sin \theta + \cos\theta - 1$.
Now observe that a unit square with angle of rotation $\theta$ (for $\theta \neq 0,\pi/2$) fails to cover any corners of $T$ if and only if its center lies in the interior of $T_\theta'$. If $\theta = 0,\pi/2$, clearly the unit square must cover a corner of $T$ no matter where inside $T$ its center lies.
To prove the reverse direction, if a unit square covers a corner of $T$, that corner lies on a side of $T_\theta$ that intersects with $T$. But then the center of the unit square is at a distance of at most $\frac{1}2$ from that side of $T_\theta$. To check the forward direction, note that ithe square $T_\theta$ can be dissected into 4 $\frac{1}2$ by $\sin\theta + \cos\theta - 1$ rectangles and 4 $\frac{1}2\cdot \frac{1}2$ squares.

But I'm not sure how to show if the dropped unit square must cover a corner of $T$ in this case.

Thus the required probability is equal to $\frac{2}{\pi} \int_0^{\frac\pi2} (\sin \theta + \cos\theta-1)^2d\theta = 2-\frac{6}{\pi}$ ($(\sin \theta + \cos\theta-1)^2$ is the probability of the square not touching any corners of $T$ provided it is rotated by an angle $\theta \in [0,\pi/2]$) .

Edit: For the current bountied question, I'm seeking a formal justification (not just a diagram or intuitive justification) for the claim that if $\theta \neq 0,\pi/2$, where $\theta$ is the (counterclockwise) angle of rotation of $T$ that is used to obtain $T_\theta$, and if the center of the unit square is in the filled square $T$ but not in the interior of $T_\theta'$, then the dropped unit square must intersect a corner of $T$.

 A: 
Can someone formally prove why the side lengths of $T_{\theta}$ are $\sin\theta+\cos\theta$?

Note that the keyword is "... circumscribe a square $T_{\theta}$ around $T$..." The intended mental picture is that $T_{\theta}$ is the biggest (and rotated) outside of the unit square $T$, and $T_{\theta}'$ is the smallest inside. The two pictures below each show a possible rotation by $\theta$.

The black "upright" square $T$ is the unit side lengths with $AB=BC=CD=DA=1$. The sides of the green $T_{\theta}$ can be calculated as, for example, $EH=AH+AE$.
\begin{align}
&AE = AB \cdot \cos\theta = \cos\theta \\
&AH=BE =  AB \cdot \sin\theta = \sin\theta \\
&\implies EH = \sin\theta+\cos\theta
\end{align}

But I'm not sure how to show if the dropped unit square must cover a corner of T in this case.

Try and see if this GeoGebra applet helps.one might need to close the panel on the right hand side (via clicking the X on top right corner) in case the view is blocked
One can move the point $E$ along a dashed semicircle to change the rotation angle $\theta$, and one can freely move the blue point $P$ that is the center of the dropped unit square (in grey, as opposed to $ABCD$ in black).
If the link above that runs the applet in a browser is unwieldy, here's a download link to run it with the app installed on your device.
