There are some problems of notation. For example, we are in effect told that $\frac{dx}{dt}=0.1$. (It is not $\frac{dt}{dx}$.)
And we are asked to find how fast the angle is changing, so we want $\frac{d\theta}{dt}$.
The relationship $\frac{x}{13}=\cos\theta$ is right. Differentiate both sides with respect to $t$. We get
$$\frac{1}{13}\cdot \frac{dx}{dt}=-\sin\theta\frac{d\theta}{dt}\tag{1}$$
(Chain Rule).
Now freeze the situation at the instant when the height is $12$. At that instant, we have $x=5$, and $\frac{dx}{dt}=0.1$, and $\sin\theta=\frac{12}{13}$.
Finally, solve for $\frac{d\theta}{dt}$. We get that at that instant $\frac{d\theta}{dt}=-\frac{0.1}{12}$ radians per second.
Well, $\theta$ is in degrees, so the derivative of $\sin(\theta)$ is not $\cos(\theta)$ but $\frac{\pi}{180}\cos(\theta)$. This is because:
$$\frac{d}{dx}[\sin(\theta)] = \cos(\theta)$$ for radian values of $\theta$ only.
To get degree values:
$\frac{d}{dx}[\sin(\theta)]$ with $\theta$ in radians is equal to $\frac{d}{dx}[\sin(\frac{\pi}{180}\cdot \theta)]$ with theta in degree values but the sine function still taking radian values. Therefore, the new derivative is:
$$\frac{d}{dx}[\sin(\frac{\pi}{180}\cdot \theta)] = \frac{\pi}{180}\cdot \cos(\frac{\pi}{180}\cdot \theta)$$
where theta is in degrees but cosine is still taking radian values. Therefore, if we change cosine and sine to take degree values we can simplify to:
$$\frac{d}{dx}[\sin(\theta)] = \frac{\pi}{180}\cdot \cos(\theta)$$ for degree values of $\theta$
Therefore, with your previous evaluation, I believe the correct answer is actually $\frac{\pi}{36}$. If I assume that as your answer, I can answer your other questions:
For your first question, we can be sure of that because $x'$ is equal to: $\frac{dx}{d\theta}$, which is read as the rate in change of $x$ with respect to $\theta$.
For your second question, $x$ is already in feet and $\theta = 60$ degrees (as stated in the question), so all you have to do is state it like this:
The rate of change of $x$ with respect to $\theta$ at $\theta$ = 60 degrees is $\frac{\pi}{36}$ feet per degree.
Best Answer
We have $(a+x)^2+y^2=x^2+(b+y)^2$ (as they are both equal to the length of the ladder). And $\tan\alpha=\frac{b+y}{x}$ and $\tan\beta=\frac{y}{a+x}$. Eliminating $x,y$ between these three equation, we get $$ a^2 \tan (\beta )+2 a b-b^2 \tan (\beta )=\tan (\alpha ) \left(-a^2+2 a b \tan (\beta )+b^2\right)\ . $$ Dividing both sides by $ab$, we get $$ (a/b) \tan (\beta )+2 -(b/a) \tan (\beta )=\tan (\alpha ) \left(-(a/b)+2 \tan (\beta )+(b/a)\right)\ . $$ Setting $a/b=\xi$, this is equivalent to $$ \xi \tan (\beta )+2 -(1/\xi) \tan (\beta )=\tan (\alpha ) \left(-\xi+2 \tan (\beta )+(1/\xi)\right)\ , $$ which can be solved for $\xi$.
EDIT: Solving for $\xi$, one obtains two solutions:$\xi=-\mathrm{cot}((\alpha+\beta)/2)$ (which is not acceptable because, for $0<\alpha+\beta<\pi$ it gives a negative value for the ratio $a/b$), and $\xi=\tan((\alpha+\beta)/2)$. Then, using a tangent of half-angle formula https://en.wikipedia.org/wiki/Tangent_half-angle_formula I can write this as $$ \xi=\frac{a}{b}=\frac{\cos\beta-\cos\alpha}{\sin\alpha-\sin\beta}\ , $$ i.e. with a minus sign difference with respect to the solution posted by the OP, which seems in fact incorrect. Take for example the case $\alpha=\pi/4$ and $\beta=\pi/6$. Then we have $x=b+y=L/\sqrt{2}$, $y=L/2$ and $a+x=(L/2)\sqrt{3}$, if $L$ is the length of the ladder. Then, $a/b=-2-\sqrt{2}+\sqrt{3}+\sqrt{6}\approx 0.76... $, while $\frac{\cos (\alpha )-\cos (\beta )}{\sin (\alpha )-\sin (\beta )}=\left(1+\sqrt{2}\right) \left(\sqrt{2}-\sqrt{3}\right)\approx -0.76...$