= Solution
For an odd prime $p$, the <Legendre symbol> $(a/p)$ is zero if $p\mid a$, one if $a$ is a nonzero square modulo $p$, and minus one otherwise. By <Gauss lemma (number theory)>, $(2/p)=(-1)^N$, where $N$ counts those least positive residues of $2,4,\ldots,p-1$ which exceed $p/2$. They correspond to integers $j>p/4$ among $1\le j\le(p-1)/2$, hence
$$
N=\frac{p-1}{2}-\left\lfloor\frac p4\right\rfloor.
$$
For $p\equiv1,3,5,7\pmod8$, its parity is respectively $0,1,1,0$, the same as that of $(p^2-1)/8$. Therefore
$$
\boxed{\left(\frac2p\right)=(-1)^{(p^2-1)/8}.}
$$
Use multiplicativity and <quadratic reciprocity> for the requested composite numerator. Since $91=7\cdot13$,
$$
\left(\frac7{167}\right)=-\left(\frac{167}7\right)=-\left(\frac{-1}7\right)=1,\qquad \left(\frac{13}{167}\right)=\left(\frac{11}{13}\right)=\left(\frac2{11}\right)=-1.
$$
The first reciprocity sign is negative because both primes are $3\pmod4$; the subsequent signs are positive because one prime is $1\pmod4$. Thus \b[$(91/167)=-1$].
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