Find the fundamental period of $f(x)=\sin\left(2m\pi\{x\}\right)$ and $g(x)=\sin\left((2m+1)\pi\{x\}\right)$, where $\{x\}$ denotes the fractional part of $x$ and $m$ is a natural number.
Since the period of $\{x\}$ is $1$ so period of $f(x)$ and $g(x)$ should be $1$ but the answers given are $\frac{1}{m}$ for $f(x)$ and $\frac{2}{2m+1}$ for $g(x)$.
It is highly confusing. I plotted their graphs and found that periods are indeed $\frac{1}{m}$ and $\frac{2}{2m+1}$
Since both functions depend on $x$ only via the fractional part $\{x\}$, it's clear that $1$ is a period, as you realized. Your main error is that because you found one period of the respective functions, you assumed it was the smallest (or fundamental) period.
$f(x)$ will make $m$ full oscillations in the interval [0,1). That means it's fundamental period will be $\frac1m$.
For $g(x)$, could there be an error in your transcription of the problem? I ask because the correct answer (for the $g(x)$ you stated) is indeed that the fundamental period is $1$.
One hint that $\frac{2}{2m+1}$ cannot be correct is that if a fundamental period exists, all periods are integer multiples of the fundamental period. But $1=\frac{2m+1}2\frac2{2m+1}$ and $\frac{2m+1}2=m+\frac12$ is not an integer.
I've plotted $g(x)$ for $m=1$:
If you look at the minima (which appear $0.5, 1.5,2.5,\ldots)$, it's clear that there can't be a perioded less than $1$.
For more higher $m$, there will always be that "M" shape around the integers ($z$) where the "old" period ends in a half cycle > 0 in the interval $[z-\frac1{2m+1},z]$ and the next period starts also with a half cycle > 0 in the interval $[z,z+\frac1{2m+1}]$.
This "M" part of the graph can not be found elsewhere, it's due to the $\{x\}$ function being discontinuous at integer values, and it isn't discontinuous at other values, so at other values $g(x)$ is "smooth" and doesn't have points where it can't be differentiated.