= Solution
For an odd <prime> $p$, the <Legendre symbol> is
$$
\left(\frac ap\right)=
\begin{cases}
0,&p\mid a,\\
1,&a\not\equiv0\pmod p\text{ and }a\text{ is a quadratic residue},\\
-1,&a\text{ is a quadratic nonresidue}.
\end{cases}
$$
<Gauss lemma> says that if $p\nmid a$ and $N$ of the least positive residues of
$$
a,2a,\ldots,\frac{p-1}{2}a
$$
exceed $p/2$, then $(a/p)=(-1)^N$.
For $a=2$, none of $2,4,\ldots,p-1$ needs reduction modulo $p$, and the terms exceeding $p/2$ are those with $j>p/4$. Hence
$$
N=\frac{p-1}{2}-\left\lfloor\frac p4\right\rfloor.
$$
Checking $p\equiv1,3,5,7\pmod8$ shows that $N$ has the same parity as $(p^2-1)/8$. Therefore
$$
\boxed{\left(\frac2p\right)=(-1)^{(p^2-1)/8}.}
$$
By <quadratic reciprocity>, and because $149\equiv1\pmod4$,
$$
\left(\frac{105}{149}\right)
=\left(\frac3{149}\right)
\left(\frac5{149}\right)
\left(\frac7{149}\right)
=\left(\frac2{3}\right)
\left(\frac4{5}\right)
\left(\frac2{7}\right).
$$
The three factors are $-1,1,1$, respectively, so
$$
\boxed{\left(\frac{105}{149}\right)=-1.}
$$
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