How would I write an acceptance-rejection algorithm for this problem

As it is written, your solution is wrong.

First and foremost, the second density function $h$ must have a typo in it, since a density function has to be non-negative, and $h(x)=kx$ takes on both negative and positive values on the interval $[-\pi/4,\pi/2]$.

I will assume from now on that the correct density function is $h(x)=k$, meaning $h$ has a uniform density, so the correct value of $k$ (such that the density integrates to $1$ over $[-\pi/4,\pi/2]$) is $k= \frac{4}{3\pi}$.

Setting that aside, notice that we are looking to generate random values that lie in the interval $[0,\pi/2]$. Anything outside of this interval should always be rejected.

So you are supposed to generate Y values according to the uniform distribution on $[-\pi/4,\pi/2]$, and then evaluate the ratio $\frac{f(Y)}{ch(Y)}$.

But $f(y)=\cos(y)$ when $y\in [0,\pi/2]$ and $f(y)=0$ everywhere else. You have to explicitly take this restriction into consideration when evaluating the ratio or you will end up accepting some values outside the interval $[0,\pi/2]$.

Secondly, $c:= \sup_y \frac{f(y)}{h(y)} = \sup_y \frac{cos(y)}{4/3\pi} = \frac{3\pi}{4}$.

So you generate a uniform variable $U$ over $[0,1]$, then generate a random variable $Y$ with density $h(y)$ (uniform density over $[-\pi/4,\pi/2]$), and then accept $Y$ if $U \leq \frac{f(Y)}{ch(Y)}$.

A brief overview of this method can be found here:

http://www.columbia.edu/~ks20/4703-Sigman/4703-07-Notes-ARM.pdf

Edit: Best thing you could do to make sure you got the right answer is to generate the variables yourself and see if you get a histogram that closely resembles the density of $f$.