The common thread between these two questions is that they both involve breaking integers down into their unique prime factor decomposition (using the Fundamental theorem of arithmetic). This allows you to find common factors in the relevant expressions and simplify to obtain solutions.
(i) When you want to find the greatest common divisor of two numbers, the simplest thing to do is to break those numbers down into their unique prime factor decomposition:
$$\begin{equation} \begin{aligned}
315 &= 3^2 \cdot 5 \cdot 7, \\[6pt]
495 &= 3^2 \cdot 5 \cdot 11. \\[6pt]
\end{aligned} \end{equation}$$
The greatest common divisor of these two numbers will consist of the shared prime factors, so it is:
$$\text{gcd}(315, 495) = 3^2 \times 5 = 45.$$
(ii) For the latter problem, you are trying to find integers $x$ and $k$ for which:
$$495 \cdot x = 6 \cdot k + 90.$$
Again, we can break the relevant numbers down into their unique prime factor decomposition:
$$3^2 \cdot 5 \cdot 11 \cdot x = 2 \cdot 3 \cdot k + 2 \cdot 3^2 \cdot 5.$$
Eliminating the common factor and grouping terms gives:
$$3 \cdot 5 \cdot 11 \cdot x = 2 \cdot (k + 3 \cdot 5).$$
So a solution is obtained by taking $k=3 \cdot 5 \cdot 10 = 150$ and $x=2$. That is:
$$990 \equiv 90 \quad \text{(mod } 6 \text{)}.$$
Instead of using the primitive root $2$, note that $7$ is a primitive root mod 11, because $7^2\equiv 5$ and $7^5\equiv -1$. So using 7-indices is a sensible move.
We list the values:
$$
\begin{array}{c|c|c}
\hline
x\pmod{11} & 5x^3\equiv 7^x\pmod{11} & x\pmod{10}
\\\hline
1 & 5&2\\
2 & 7&1\\
3 & 3&4\\
4 & 1&0\\
5 & 9&8\\
6 & 2&3\\
7 &10&5\\
8 & 8&9\\
9 & 4&6\\
10& 6&7\\\hline
\end{array}
$$
So combining the outer columns, you know ten solutions $x\pmod{110}$ and you know $x\equiv 2\pmod{3}$. Hence you can write down the ten solutions mod 330, and determine the solutions between 0 and 110.
Best Answer
Hint $\,\bmod 9\,$ invertibles have form $\,2^{\large n}$ so $\,2^{\large n} x\equiv 2 \iff x \equiv 2^{\large\:\! 7-n},\,\ n = 0,1,2,\ldots,5 $
Example $\ $ For $\,n = 2\,$ the above says that $\, 2^{\large 2} x\equiv 2\iff x\equiv 2^{\large 5}\equiv 5.\,$ Indeed $\,2^{\large 2} 5\equiv 2\,\color{#c00}\checkmark$
Remark $\ $ Since $\,-7\equiv 2\,$ is invertible $\bmod 9\,$ so too is its factor $\,a := 2n\!+\!1.\,$ Or, more explicitly, $\,ax\equiv 2\,\iff ax\,2^{-1}\equiv 1\iff a^{-1}\equiv 2^{-1}x\equiv 5x$
That every invertible has form $\,2^{\large n}$ follows because $\,2\,$ is a primitive root $\bmod 3^{\large 2}\,$ (generally a p.r. $\,g \bmod p\,$ persists as a p.r. $\bmod p^k\,$ except if $\, g^{\large p-1}\!\equiv 1\pmod{\!p^2};\,$ where instead $\,g\! +\! p\,$ works).
Directly: $\,a\,$ invertible $\!\bmod 9\iff a\,$ invertible $\!\bmod 3\iff a = \pm1 + 3j,\,$ so
$\!\bmod \color{#c00}9\!:\,\ ax = (\pm1 + 3j)x \equiv 2\iff x\equiv \dfrac{2}{\pm1 + 3j} \equiv \dfrac{2(1-\color{#c00}9j^2)}{\pm1 + 3j\ \ } \equiv 2(\pm1 -3j)$
Example $\ \ \ \ \ \ a = 1+3\iff x \equiv 2(1-3)\equiv 5,\,$ same as above