Below is a proof that works in any domain, using the universal definitions of GCD, LCM.
Theorem $\rm\quad (a,b)\ =\ ab/[a,b] \;\;$ if $\;\rm\ [a,b] \;$ exists.
Proof: $\rm\ \ \ \ \ \ d\mid a,b \iff a,b\mid ab/d \iff [a,b]\mid ab/d \iff d\mid ab/[a,b] $
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 $\rm\:\ n\:\equiv\: 2 + 7\cdot 3\equiv 23\pmod{7\cdot 5}$
Best Answer
Notation: $\text{HCF}$ is denoted below as $\text{gcd}$.
Assume you have two fractions $\frac{a}{b},\frac{c}{d}$ reduced to lowest terms. Let
$$\begin{eqnarray*} a &=&\underset{i}{\prod } p_{i}^{e_{i}(a)},\qquad b=\underset{i}{\prod } p_{i}^{e_{i}(b)}, \\ c &=&\underset{i}{\prod } p_{i}^{e_{i}(c)},\qquad d=\underset{i}{\prod } p_{i}^{e_{i}(d)}. \end{eqnarray*}$$
be the prime factorizations of the integers $a,b,c$ and $d$. Then
$$\frac{\underset{i}{\prod }\ p_{i}^{\max \left( e_{i}(a),e_{i}(c)\right) }}{% \prod_{i}\ p_{i}^{\min \left( e_{i}(b),e_{i}(d)\right) }}$$
is a fraction which is a common multiple of $\frac{a}{b},\frac{c}{d}$. It is the least one because by the properties of the $\text{lcm}$ and $\gcd $ of two integers, $\prod_{i}\ p_{i}^{\max \left( e_{i}(a),e_{i}(c)\right) }$ is the least common multiple of the numerators and $\prod_{i}\ p_{i}^{\min \left( e_{i}(b),e_{i}(d)\right) }$ is the greatest common divisor of the denominators. Hence
$$\begin{eqnarray*} \text{lcm}\left( \frac{a}{b},\frac{c}{d}\right) &=&\text{lcm}\left( \frac{{\prod_{i}\ p_{i}^{e_{i}(a)}}}{\prod_{i}\ p_{i}^{e_{i}(b)}},\frac{% \prod_{i}\ p_{i}^{e_{i}(c)}}{\prod_{i}\ p_{i}^{e_{i}(d)}}\right)=\frac{\prod_{i}\ p_{i}^{\max \left( e_{i}(a),e_{i}(c)\right) }}{% \prod_{i}\ p_{i}^{\min \left( e_{i}(b),e_{i}(d)\right) }}=\frac{\text{lcm}(a,c)}{\gcd (b,d)}.\quad(1) \end{eqnarray*}$$
Similarly
$$\begin{eqnarray*} \gcd \left( \frac{a}{b},\frac{c}{d}\right) =\gcd \left( \frac{{\prod_{i}\ p_{i}^{e_{i}(a)}}}{\prod_{i}p_{i}^{e_{i}(b)}},\frac{% \prod_{i\ }p_{i}^{e_{i}(c)}}{\prod_{i}p_{i}^{e_{i}(d)}}\right) =\frac{\prod_{i}\ p_{i}^{\min \left( e_{i}(a),e_{i}(c)\right) }}{% \prod_{i}\ p_{i}^{\max \left( e_{i}(b),e_{i}(d)\right) }} =\frac{\gcd (a,c)}{\text{lcm}(b,d)}.\quad(2) \end{eqnarray*}$$
The repeated application of these relations generalizes the result to any finite number of fractions.