Lets assume $d$ is a natural number which makes $(n+1)/(n+3)$ reducible, then $d|n+1$ and $d|n+3$.
$d|[n+3-(n+1)] = d|2$ which means $d=1$ or $d=2$.
$n+1$ and $n+3$ must be divisible by $2$ so all natural numbers of the form $2n+1$ will work.
Now $\text{gcd}(n+1,n+3) = 1$ so shouldn't this fraction be irreducible for any $n$? ($n$ natural number)