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 $\rm\:\ n\:\equiv\: 2 + 7\cdot 3\equiv 23\pmod{7\cdot 5}$
(1) $12$ has to be part of both numbers - since it is the GCD.
(2) In order to get the LCM as $168$ we got to distribute the remaining $14$ among these two numbers, being careful that the distribution has no common factors.
(3) The factors of $14$ are $1,2,7,14$. So we take the first number as $12.1$, $12.2$, $12.7$, $12.14$ and the second number as $12.14$,$12.7$,$12.2$ and $12.1$ respectively
(4) Given that we exclude the extremities, we have only one value up to symmetry.
Finally, we have $\color{blue}{(A,B) \in \{(24,84), (84,24)\}}$
Some Notes
This problem becomes a little interesting if the LCM given was say $336$. Now you have to be more careful in distributing the $28$. It cannot be split as $2$ and $2$, otherwise it will affect the GCD. So you have to take the $4$ factor completely.
lab bhattacharjee has of of course a more formal approach.
Best Answer
Both numbes must be multiples of $6$ and at least one must be a multiple of $5$ (because $5$ is a prime dividing $15$). So the first attempt is $a=30$ and $b=6$ and works