With different types of equations come different techniques, and several different ones will be used in the following, but the key idea is usually to look for symmetry.
We can actually solve the first equation $2^{\cos x}=|\sin x|$ extremely quickly given those answer choices. We note that $|\sin x|$ has period $\pi$ and that $2^{\cos x}$ has period $2\pi$, but we also have $\cos(x)=\cos(-x)$, so actually the number of solutions on the interval $[0,\pi]$ is the same as the number of solutions on any $[n\pi,(n+1)\pi]$ interval.
Now, we have $7$ such intervals in our question, so the solution must be a multiple of $7$ (well, you do first have to check the edge case that there are no solutions on the boundary of the intervals, but that's easy to do). Since the only multiple of $7$ among our answer choices is $14$, we have our solution.
If we didn't have the answer choices, this would be much more annoying and we would need some more technical arguments to say that there are exactly $2$ solutions on the interval $[0,\pi]$ (which would probably start by noticing that there is a solution at $\frac{\pi}{2}$).
As for the other queries:
Question 1: $\sin\{x\}=\cos\{x\}$ on $[0,2\pi]$.
For this one, all you have to do is note that $\{x\}$ has period $1$, so this function on the interval $[0,1)$ is exactly the same as on any other interval $[n,n+1]$. Now, we note that the only solution to $\sin(x)=\cos(x)$ on the interval $[0,1)$ is $\frac{\pi}{4}$ (a well known trig result). Hence the only solutions we have are in the form $n+\frac{\pi}{4}$ where $n$ is an integer. This immediately gives us $6$ solutions on the specified interval, with the only question left as whether $6+\frac{\pi}{4}\in[6,2\pi]$. However, we know $2\pi<6.3$ and $\frac{\pi}{4}>\frac{3}{4}=.75$, so we conclude there is no solution on the $[6,2\pi]$ interval, which means we have $6$ solutions in total.
Question 2: $x+2\tan x = \dfrac{\pi}{2}$ on $[0, 2\pi]$.
First note that this function is always increasing (except at the discontinuities where $\cos(x)=0$). Now $\tan(x)$ starts each interval $(n\pi-\frac{\pi}{2},n\pi+\frac{\pi}{2})$ at $-\infty$ and ends each interval at $\infty$. Hence there is exactly $1$ solution on each interval (exactly one place where this function takes on the value $\frac{\pi}{2}$). For our situation, we contain just $1$ complete interval: $[\frac{\pi}{2},\frac{3\pi}{2}]$, so we get $1$ solution from that. Now for the half interval $[0,\frac{\pi}{2})$, we have exactly one solution for that as well because $0+2\tan(0)=0<\frac{\pi}{2}$ and the function is always increasing (to infinity). For the other half interval $(\frac{3\pi}{2},2\pi]$, we have again $1$ solution by the same reasoning: we start at $-\infty$ and get to $2\pi+2\tan(2\pi)=2\pi>\frac{\pi}{2}$, so somewhere in there we contain exactly $1$ solution. This gives us $3$ solutions in total.
Note the key idea in this was to apply the fact that on every interval we have an always increasing function.
Question 3: $\sin x = \dfrac{|x|}{10}$.
We note that $\sin x\le 1$, so the only possible solutions are in the range $[-10,10]$. Now, since $\frac{|x|}{10}$ is just a straight line starting at $0$ and going to $1$ on the interval $[0,10]$, every time that $\sin x$ has a peak on the interval, we get two solutions as the line cuts across. Since $\sin$ has peaks at $x=\frac{\pi}{2}+2\pi n$, we have just $2$ peaks on the interval $[0,10]$, so we have $4$ solutions on this side. Similarly we just need to count the peaks for the interval $[-10,0)$ and we get just one more peak, so there are $4+2=6$ solutions in total.
Techniques: The techniques used in these are mainly breaking the function up into periodic or otherwise symmetric pieces and then applying arguments from there.
