Period of $f(2x+3)+f(2x+7)=2$

I found this tricky and I may have got it wrong, but I think that you are correct and the book is not.

If $f(2x+3)+f(2x+7)=2$ for all real $x$, then we can take $x=(y-3)/2$ to give $$f(y)+f(y+4)=2\tag{$*$}$$ and then, essentially following your argument, $$f(y)=f(y+8)$$ for all real $y$. Now suppose that $f$ has period $t$: recall that this means $t>0$ and $f(y)=f(y+t)$ for all real $y$, and that no smaller positive $t$ has the same property. Suppose that there is an integer $k$ such that $$kt<8<(k+1)t\ ;$$ this can be rewritten as $$8=kt+t'\quad\hbox{with}\quad 0<t'<t\ .$$ Then we have $$f(y)=f(y+8)=f(y+t'+kt)=f(y+t')\quad\hbox{for all $y$},$$ which contradicts the fact that $f$ has period $t$. Therefore we must have $8=kt$ for some positive integer $k$, that is, $t=8/k$.

Now $k$ cannot be even, for if $k=2m$ then for all $y$ we have $f(y)=f(y+mt)=f(y+4)$; in conjunction with $(*)$ this shows that $f(y)=1$ for all $y$, which is not (strictly speaking) periodic.

However it is possible that the period could be $t=8/k$ with $k$ odd. For example, take $$f(y)=1+\sin\Bigl(\frac{2\pi y}{t}\Bigr)\ .$$ Then, as is well known, $f$ has period $t$; also $$f(y)+f(y+4) =1+\sin\Bigl(\frac{2\pi y}{t}\Bigr)+1+\sin\Bigl(\frac{2\pi y}{t}+k\pi\Bigr)=2$$ for all $y$, as required.

"How can I be sure that the t so found is the least?"

Answer : by finding an example where $8$ is the smallest period. For example, take $f$ such that $f(x)=\frac{x-8k}{2}$ when $x\in[8k,8k+4[$ and $f(x)=\frac{8k+8-x}{2}$ when $x\in[8k+4,8k+8[$, for every $k\in{\mathbb Z}$.

This creates a "sawtooth" function and only multiples of $8$ are periods.