= Solution
The <Wilson theorem> says that $(p-1)!\equiv-1\pmod p$ for a <prime number> $p$. Thus $18\cdot17!\equiv-1\pmod{19}$, and since $18\equiv-1$, we get $17!\equiv1$. The <Fermat little theorem> says $b^{p-1}\equiv1\pmod p$ when $p$ does not divide $b$. Applying it to $3$ gives $3^{18}\equiv1$, so $3^{16}$ is the inverse of $9$ modulo $19$. Since $9\cdot17=153\equiv1$, the required residue is
$$
\boxed{a=17.}
$$
This is the unique representative in $\{1,\ldots,19\}$.
For the simultaneous <integer congruences>, the modulus-$2$ condition is already implied by $x\equiv3\pmod4$. Write $x=3+4k$; the modulus-$3$ condition becomes $k\equiv1\pmod3$, hence $x=7+12t$. The modulus-$5$ condition then gives $2+2t\equiv4$, or $t\equiv1\pmod5$. Therefore
$$
\boxed{x=19+60j,\qquad j\in\mathbb Z.}
$$
Every such integer satisfies all four <integer congruences>; reversing the substitutions shows that there are no others. The uniqueness modulo $60$ also follows from the <Chinese remainder theorem> applied to the coprime moduli $3,4,5$.
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