One method is simply the Euclidean algorithm:
\begin{align*}
31 &= 4(7) + 3\\\
7 &= 2(3) + 1.
\end{align*}
So $ 1 = 7 - 2(3) = 7 - 2(31 - 4(7)) = 9(7) - 2(31)$. Viewing the equation $1 = 9(7) -2(31)$ modulo $31$ gives $ 1 \equiv 9(7)\pmod{31}$, so the multiplicative inverse of $7$ modulo $31$ is $9$. This works in any situation where you want to find the multiplicative inverse of $a$ modulo $m$, provided of course that such a thing exists (i.e., $\gcd(a,m) = 1$). The Euclidean Algorithm gives you a constructive way of finding $r$ and $s$ such that $ar+ms = \gcd(a,m)$, but if you manage to find $r$ and $s$ some other way, that will do it too. As soon as you have $ar+ms=1$, that means that $r$ is the modular inverse of $a$ modulo $m$, since the equation immediately yields $ar\equiv 1 \pmod{m}$.
Another method is to play with fractions Gauss's method:
$$\frac{1}{7} = \frac{1\times 5}{7\times 5} = \frac{5}{35} = \frac{5}{4} = \frac{5\times 8}{4\times 8} = \frac{40}{32} = \frac{9}{1}.$$
Here, you reduce modulo $31$ where appropriate, and the only thing to be careful of is that you should only multiply and divide by things relatively prime to the modulus. Here, since $31$ is prime, this is easy. At each step, I just multiplied by the smallest number that would yield a reduction; so first I multiplied by $5$ because that's the smallest multiple of $7$ that is larger than $32$, and later I multiplied by $8$ because it was the smallest multiple of $4$ that is larger than $32$. Added: As Bill notes, the method may fail for composite moduli.
Both of the above methods work for general modulus, not just for a prime modulus (though Method 2 may fail in that situation); of course, you can only find multiplicative inverses if the number is relatively prime to the modulus.
Update. Yes, your method for general linear congruences is the standard one. We have a very straightforward method for solving congruences of the form $$ax \equiv b\pmod{m},$$
namely, it has solutions if and only if $\gcd(a,m)|b$, in which case it has exactly $\gcd(a,m)$ solutions modulo $m$.
To solve such equations, you first consider the case with $\gcd(a,m)=1$, in which case $ax\equiv b\pmod{m}$ is solved either by finding the multiplicative inverse of $a$ modulo $m$, or as I did in method $2$ above looking at $\frac{b}{a}$.
Once you know how to solve them in the case where $\gcd(a,m)=1$, you can take the general case of $\gcd(a,m) = d$, and from
$$ax\equiv b\pmod{m}$$
go to
$$\frac{a}{d}x \equiv \frac{b}{d}\pmod{\frac{m}{d}},$$
to get the unique solution $\mathbf{x}_0$. Once you have that unique solution, you get all solutions to the original congruence by considering
$$\mathbf{x}_0,\quad \mathbf{x}_0 + \frac{m}{d},\quad \mathbf{x}_0 + \frac{2m}{d},\quad\ldots, \mathbf{x}_0 + \frac{(d-1)m}{d}.$$
The $d^2 \equiv -1 \pmod{p}$ part is wrong. Presumably, the author mixed some parts of the $p \equiv 1 \pmod{8}$ case in, where one computes $d = y^{(p-1)/8}$.
For $p \equiv 5 \pmod{8}$, we always have $1 = \left(\frac{y}{p}\right) \equiv d^2 \pmod{p}$.
So if $d \equiv 1 \pmod{p}$, we choose $r \equiv y^{(p+3)/8}\pmod{p}$, and that yields
$$r^2 \equiv y^{(p+3)/4} \equiv y\cdot y^{(p-1)/4} \equiv y\cdot d \equiv y \pmod{p}.$$
For $d \equiv -1\pmod{p}$, choosing $r \equiv y^{(p+3)/8} \pmod{p}$ would lead to $r^2 \equiv y\cdot d \equiv -y \pmod{p}$ by the same computation as above, so we mix a square root of $-1$ in to obtain a square root of $y$. For $p \equiv 5 \pmod{8}$, we have $\left(\frac2p\right) = -1$, so $2^{(p-1)/4}$ is a square root of $-1$ modulo $p$. The author just wrote it in an obscure way, we have $2\cdot 4^{(p-5)/8} = 2\cdot 2^{(p-5)/4} = 2^{(p-1)/4}$.
Best Answer
$$7^2=49=50-1\implies7^4=(50-1)^2=50^2-2\cdot50+1^2\equiv1\pmod{100}$$
As $\displaystyle30\equiv2\pmod4,7^{30}\equiv7^2\pmod{100}$
$\displaystyle7\equiv-1\pmod4\implies7^{30}\equiv(-1)^{30}\implies7^{30}\equiv1\pmod4\ \ \ \ (1)$
$\displaystyle7^2=49\equiv-1\pmod{25}\implies7^{30}=(-1)^{15}\implies7^{30}\equiv-1\pmod{25}\ \ \ \ (2)$
Apply Chinese Remainder Theorem on $(1),(2)$