How: For how to do it, look for the Extended Euclidean Algorithm. It is not particularly hard to carry out, but a proper description would be quite lengthy. The Extended Euclidean Algorithm is also described in most books on Elementary Number Theory. There are many computer implementations, but for $a$ and $c$ of modst size, the procedure can be easily carried out by hand.
If $d$ is the greatest common divisor of $a$ and $c$, the Extended euclidean algorithm produces integers $x$ and $y$ such that $ax+d=cy$. But from that one can solve your more general problem.
Existence of solution: Certainly there do not always exist such $m$ and $n$. For example, let $a=2$, $b=1$, $c=2$. Then $an+b$ (that is, $2n+1$) is always odd, and $cm$ (that is, 2m$) is always even, so they can never be equal.
In general, if the greatest common divisor of $a$ and $c$ divides $b$, there always is a solution. If the greatest common divisor of $a$ and $c$ does not divide $b$, there is no solution.
Infinitely many solutions: If there is a solution $(n_0,m_0)$, then there are infinitely many solutions. For suppose that $an_0+b=cm_0$. Then for any integer $t$, we have
$$a(n_0 +ct)+b=c(n_0+at).$$
(Just expand. There will be a term $act$ on the left, and a term $cat$ on the right, and they cancel.)
So if $(n_0,m_0)$ is a solution, then so is $(n_0+ct, m_0+at)$ for any positive integer $t$.
They are the only possible solution. Proof is as follows:
Suppose that $d = gcd(x,y)$ and $x=d ~ x0$, $y=d ~y_0$ where $x_0$ and $y_0$ are co-prime.
Substituting in the original equation we get
$$ 1/x + 1/y + 1/z=1 \Rightarrow -d\,x_0\,y_0\,z+y_0\,z+x_0\,z+d\,x_0\,y_0 =0$$
Solving for $d$:
$$d=\frac{\left( y_0+x_0\right) \,z}{x_0\,y_0\,\left( z-1\right) }$$
Since $x_0$ and $y_0$ are co-prime, for $d$ to be an integer, $x_0\,y_0$ should divide $z$.
Hence we require
$$ z= k ~x_0 ~y_0$$
Substituting in the equation for $d$ and solving for $k$:
$$ k=\frac{d}{d\,x_0\,y_0-y_0-x_0}$$
This shows that $d$ is a multiple of $k$. Let
$$ d= \mu k$$. Then
$$k=\frac{k\,\mu}{k\,\mu\,x_0\,y_0-y_0-x_0}$$
Solving for $k$:
$$k=\frac{1}{x_0\,y_0}+\frac{1}{\mu\,y_0}+\frac{1}{\mu\,x_0}$$
This implies that $1 \le x_0 \le 3$, $1 \le y_0 \le 3$, $1 \le \mu\le 3$
It is possible to eliminate some of the 27 possible values since $k$ has to be an integers this will result in 12 possible values for $(x_0,y_0,\mu)$ and two of the solutions are repetitions giving the 10 solutions mentioned in the problem.
Best Answer
Since @Hecke has done the first one, I shall only provide the solutions for 2 and 3.
2. $p^5-q^5>p^3-q^5=(p+q)^2 \geq 0$ so $p>q$. Thus $\gcd(p, q)=\gcd(q, p+q)=\gcd(p, p+q)=1$.
Taking $\pmod{p}$, we get $p \mid q^5+q^2=q^2(q+1)(q^2-q+1)$, so $p \mid (q+1)$ or $p \mid (q^2-q+1)$.
If $p \mid q+1$, then since $p>q$, we have $p=q+1$. Now $$0<(2q+1)^2=(q+1)^3-q^5 \leq (\frac{3}{2}q)^3-q^5=q^3(\frac{27}{8}-q^2)<0$$ since $q^2 \geq 4>\frac{27}{8}$. We thus get a contradiction.
Therefore, $p \mid (q^2-q+1)$, and thus $p \leq q^2-q+1$.
Taking $\pmod{p+q}$, we get $(p+q) \mid -q^3-q^5=-q^3(1+q^2)$. Since $\gcd(q, p+q)=1$, we get $(p+q) \mid (q^2+1)$.
Now $p \mid ((q^2-q+1)-p), (p+q) \mid ((q^2-q+1)-p)$, so since $\gcd(p, p+q)=1$, $p(p+q) \mid ((q^2-q+1)-p)$.
If $p<q^2-q+1$, then we must have $((q^2-q+1)-p) \geq p(p+q)$.
But then $p^2+1>q^2+1 \geq (p+1)(p+q)>(p+1)p$, a contradiction.
Thus $p=q^2-q+1$, and we get $(q^2-q+1)^3-q^5=(q^2+1)^2$.
Now taking $\pmod{q-1}$, we get $0 \equiv 1-1 \equiv (q^2-q+1)^3-q^5 \equiv (q^2+1)^2 \equiv 2^2 \pmod{q-1}$
Thus $(q-1) \mid 4$, and so $q=2, 3, 5$.
If $q=2$, then $-5=3^3-2^5=5^2=25$, a contradiction.
If $q=3$, then $100=7^3-3^5=10^2=100$, and we get a solution $(p, q)=(7, 3)$.
If $q=5$, then $6136=9261-3125=21^3-5^5=26^2=676$, a contradiction.
Thus the only solution is $(p, q)=(7, 3)$.
3. I don't really understand the meaning of "all both positive or negative integers". Do you mean $x, y, z$ are all positive integers or all negative integers, or do you mean that $x, y, z$ are all integers, and $x, y$ are both positive or both negative? If it is the former, note that $z>0$, so then $x, y>0$. If it is the latter, which in my opinion seems more likely, better phrasing is needed.
$\frac{13}{x^2}+\frac{1996}{y^2}=\frac{z}{1997}$. Let $\gcd(x, y)=d, d>0, x=dx', y=dy'$, then $1997(13y'^2+1996x'^2)=zd^2x'^2y'^2$
Now $x'^2 \mid 1997(13)y'^2$ so since $\gcd(x', y')=1$, $x'^2 \mid 1997(13)$. $(1997)(13)$ is squarefree, so $x'=\pm 1$. Similarly, $y'^2 \mid (1997)(1996)x'^2$, so $y'^2 \mid (1997)(1996)=1997(499)(2^2)$. $1997(499)$ is squarefree, so $y'= \pm 1, \pm 2$
If $y'=\pm 1$, then $zd^2=1997(13+1996)=1997(41)(7^2)$, so $d \mid 7$.
If $d=1$, we get $z=1997(41)(7^2)=4011973$. We thus get the solutions $(x, y, z)=(\pm 1, \pm 1, 4011973)$.
If $d=7$, we get $z=1997(41)=81877$. We thus get $(x, y, z)=(\pm 7, \pm 7, 81877)$.
If $y'=\pm 2$, then $zd^2=1997(13 \cdot 4+1996)=1997(2048)=1997(2^{11})$. Thus $d \mid 2^5$.
Let $d=2^a, a=0, 1, 2, 3, 4, 5$, then $z=1997(2^{11-2a})$, and we get $(x, y, z)=(\pm 2^a, \pm 2^{a+1}, 1997(2^{11-2a}))$.
In conclusion, the solutions are $(x, y, z)=(\pm 1, \pm 1, 4011973), (\pm 7, \pm 7, 81877)$, or $(\pm 2^a, \pm 2^{a+1}, 1997(2^{11-2a}))$, where $a=0, 1, 2, 3, 4, 5$.