If $q\equiv 3 \pmod 4$ then is it true ${q+1 \choose 2}$ is semiprime if and only if $q=3$? 
Question: If $q\equiv 3 \pmod 4$ then is it true ${q+1 \choose 2}$ is semiprime if and only if
  $q=3.$

I am certain the answer to the question is yes, and I believe I have worked out a solution - although I am not certain it is correct; whence the solution-verification tag. I feel it is ok to also ask if there are solutions that are "deeper" - maybe using better applications of modular arithmetic or making use of group theory ?

Solution To Question: Since $q\equiv 3 \pmod 4$ then there exist a positive number $m\in\mathbb{N}$ such that $q=4m+3$. Then after substitution, ${q+1 \choose 2}={4m+4 \choose 2}=\frac{(4m+4)(4m+3)}{2}=(2m+2)(4m+3);$ which has even parity. In particular if $p$ is a prime number I can write $(2m+2)(4m+3)=2p.$ Division by two yields $\left(\frac{2m+2}{2}\right)\left(4m+3\right)=p.$ Recall a prime number has as divisors the numbers $1$ and itself. Note that $4m+3\neq 1$ for positive $m$ and so $\frac{2m+2}{2}=1;$ which implies $m=0$ in which case $4m+3=3.$
On the other hand if ${q+1 \choose 2}=6$ then  explicitly $\frac{q(q+1)}{2}=6$ and so $q^2+q-12=0.$ Equivalently $(q-3)(q+4)=12;$ which has roots $q=3$ or $q=-4.$ We require the positive root $q=3.$ This completes the solution to the question. $\blacksquare$ 


If $q \equiv 3 \mod 4$, $q+1$ is divisible by $4$.  Say $q+1=4k$ Then 
$$ {q+1 \choose 2} = \frac{(q+1)q}{2} = 2 k (4k-1)$$
This is divisible by $2$, and by any primes that divide $k$ or $4k-1$.  That makes at least three (not necessarily distinct) primes, unless either $k = 1$ or $4k-1 = 1$ (the latter being impossible).
