Thanks to Julian from British School Manila, Nathan from Outwood Academy Newbold and Joshua from Dubai British school for submitting solutions to this problem.
(a) Julian noticed the following:
Since $y=\sin(x)$ touches the lines $y=-1$ and $y=1$, all we need to do is translate the graph upwards by 2 units.
So a possible solution is $y=\sin(x)+2$.
(b) Nathan and Joshua both noticed:
Since $y=\cos(x)$ crosses the $x$-axis at $\frac{\pi}{2}$ and $-\frac{\pi}{2}$, we can make it cross at 1 and -1 if we stretch the graph by a factor of $\frac{2}{\pi}$ in the $x$-direction.
So a possible solution is $y=\cos(\frac{\pi}{2} x)$.
(c) Julian and Nathan both sent us solutions for this part. Here is Julian's solution:
Since $y=\tan(x)$ already has $x=\frac{\pi}{2}$ as an asymptote, we only need to translate the graph downwards by $\tan(\frac{\pi}{3})=\sqrt{3}$ to make it pass through the point $(\frac{\pi}{3},0)$.
So a possible solution is $y=\tan(x)-\sqrt{3}$.
Nathan had a different idea:
The first step is to start by make the distance between the first $x$-intercept at the origin and the first asymptote be $\frac{\pi}{6}$ by stretching the graph by a scale factor of $\frac{1}{3}$ in the $x$-direction, and then translate the graph by $\frac{\pi}{3}$ to the right to move the $x$-intercept at the origin to $x=\frac{\pi}{3}$. This also moves the asymptote back
to $x=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}$.
This leads to another possible solution of $y=\tan(3(x-\frac{\pi}{3}))=\tan(3x-\pi)$.
Thomas from BHASVIC Sixth Form College Brighton has worked towards finding conditions that would specify unique solutions to these problems:
a) The general solution to this problem is $y=2+ \sin( ax+b )$. Fixing a point $1 < y_0 < 3$ at $x=0$ and fixing a gradient $ \frac{dy}{dx} $ here at $(0, y_0)$ is sufficient to specify a unique solution since only one point in each period can go through this point and have this gradient.
b) The general solution to this problem is $y=a \cos( \frac{(2n+1) \pi x}{2})$ for any integer $n$. For uniqueness we could fix the maximum value attained by the graph and the number of periods in between $x=-1$ and $x=1$.
c) For a unique solution, we only need to specify the gradient at, say, $x=\frac{ \pi }{3}$.
However, there is more than one way of making these solutions unique.