Hint $\rm\ \ x,y,z\:|\:a\!+\!D \iff m = lcm(x,y,z)\:|\:a\!+\!D.\:$ Thus $\rm\:a\equiv -D\pmod m,\:$ which has least natural representative $\rm\:a = m\!-\!D\:$ if $\rm\:0\le D < m.$
Thus, in your example you have that $\rm\:a\:$ is congruent to a constant value mod all moduli, i.e. $\rm\:a \equiv -D \equiv -2\:\ mod\ 3,4,5,\:$ i.e. $\rm\:3,4,5\:|\:a+2\:$ $\Rightarrow$ $\rm\:60 = lcm(3,4,5)\:|\:a + 2,\:$ therefore $\rm\: a = -2 + 60\:\!k,\:$ with least natural solution $\rm\:a = 58.$
You can find much further discussion of this constant-case of CRT (Chinese Remainder) in many of my prior posts, e.g. here.
The second example isn't a constant case of CRT, but one can use an easy form of CRT, viz.
Theorem (Easy CRT) $\rm\ \ $ If $\rm\ p,\:q\:$ are coprime integers then $\rm\ p^{-1}\ $ exists $\rm\ (mod\ q)\ \ $ and
$\rm\displaystyle\quad\quad\quad\quad\quad \begin{eqnarray}\rm n&\equiv&\rm\ a\:\ (mod\ p) \\
\rm n&\equiv&\rm\ b\:\ (mod\ q)\end{eqnarray} \ \iff\ \ n\ \equiv\ a + p\ \bigg[\frac{b-a}{p}\ mod\ q\:\bigg]\ \ (mod\ p\:\!q)$
Proof $\rm\ (\Leftarrow)\ \ \ mod\ p\!:\:\ n\equiv a + p\ (\cdots)\equiv a\:,\ $ and $\rm\ mod\ q\!:\:\ n\equiv a + (b-a)\ p/p \equiv b\:.$
$\rm\ (\Rightarrow)\ \ $ The solution is unique $\rm\ (mod\ p\!\:q)\ $ since if $\rm\ x',\:x\ $ are solutions then $\rm\ x'\equiv x\ $ mod $\rm\:p,q\:$ therefore $\rm\ p,\:q\ |\ x'-x\ \Rightarrow\ p\!\:q\ |\ x'-x\ \ $ since $\rm\ \:p,\:q\:$ coprime $\rm\:\Rightarrow\ lcm(p,q) = p\!\:q\:.\quad$ QED
Applying this to your example we find
$\rm\displaystyle\quad\quad\quad\quad\quad \begin{eqnarray}\rm n&\equiv&\rm\ 2\:\ (mod\ 7) \\
\rm n&\equiv&\rm\ 3\:\ (mod\ 5)\end{eqnarray} \ \iff\ \ n\ \equiv\ 2 + 7\ \bigg[\frac{3-2}{7}\ mod\ 5\:\bigg]\ \ (mod\ 7\cdot 5)$
But $\rm\displaystyle\ mod\ 5\!:\ \frac{1}{7} \equiv \frac{6}2\equiv 3,\: $ therefore $\rm\:\ n\:\equiv\: 2 + 7\cdot 3\equiv 23\pmod{7\cdot 5}$
The $\textrm{mod}$ notation is not an operator that gives you a single output.
Don't read it as if it was computer code. I would only read the $\textrm{mod}$ notation in English. $\text{mod}\;n$ means the preceding equation/formula is under modulus $n$. In the context of integer numbers, we consider each integer number by its Euclidean remainder when divided by $n$.
$\equiv$ is a notation for equivalence relations. Please read its introductions on the internet.
For example, we say 13 is equivalent (or congruent, some even say equal, depending on the context) to 23 modulus 10, denoted by $13\equiv23\;\textrm{mod}\;10$ because they have the same remainder when divided by 10, which is 3. Negative number is usually read "in reverse". For example, $13\equiv-7\;\textrm{mod}\;10$.
As for your puzzle at the beginning, you don't have to use Chinese Remainder Theorem. I think Chinese Remainder Theorem is a bit of a leap when you are just learning the mod notation. Try solving the puzzle yourself without the theorem and share where you are first.
Best Answer
List the numbers that give remainder $1$ on division by $5$, looking for remainder $2$ on division by $3$. Start listing: $1$, $6$, $11$. Bingo!
Remark: This is probably the quickest way to solve the GRE problem. An alternative is to list the numbers that leave remainder $2$ on division by $3$, and look for remainder $1$ on division by $5$. That is a little slower: things increase faster if we count by $5$'s than if we count by $3$'s.
As to elegance, that requires thinking, and the GRE is not about that. On to the next question.