I think both congruences are essentially trivial: As $\varphi(4n+3)$ is even and $(4n+3)-1$ is $2$ times an odd number, the divisors which are counted come in pairs $t,2t$, where $t$ is odd.

Similarly for the second congruence: $(8n+5)-1$ is $4$ times an odd number, so the divisors come as triples $t,2t,4t$ for certain odd $t$'s.

So more generally, if $n=2^mu+1$ for odd $m\ge1$, then the number of divisors of $n-1$ which don't divide $\varphi(n)$ is divisible by $m+1$.

I think both congruences are essentially trivial: As $\varphi(4n+3)$ is even and $(4n+3)-1$ is $2$ times an odd number, the divisors which are counted come in pairs $t,2t$, where $t$ is odd.

Similarly for the second congruence: $(8n+5)-1$ is $4$ times an odd number, while $\varphi(8n+5)$ is divisible by $4$ (for otherwise $8n+5=p^m$ for $p\equiv3\pmod{4}$ which cannot hold), so the divisors come as triples $t,2t,4t$ for certain odd $t$'s.

I think both congruences are essentially trivial: As $\varphi(4n+3)$ is even and $(4n+3)-1$ is $2$ times an odd number, the divisors which are counted come in pairs $t,2t$, where $t$ is odd.

Similarly for the second congruence: $(8n+5)-1$ is $4$ times an odd number, so the divisors come as triples $t,2t,4t$ for certain odd $t$'s.

I think both congruences are essentially trivial: As $\varphi(4n+3)$ is even and $(4n+3)-1$ is $2$ times an odd number, the divisors which are counted come in pairs $t,2t$, where $t$ is odd.

Similarly for the second congruence: $(8n+5)-1$ is $4$ times an odd number, so the divisors come as triples $t,2t,4t$ for certain odd $t$'s.

So more generally, if $n=2^mu+1$ for odd $m\ge1$, then the number of divisors of $n-1$ which don't divide $\varphi(n)$ is divisible by $m+1$.