I'm not going to do them all for you, as I think these are valuable exercises. I will do the first and the third.
For the first one, I would split it into two cases: $i < j - 1$ or $i > j - 1$. So, my first two rules would be,
\begin{align*}
S &\to L \\
S &\to M
\end{align*}
standing for "Less" and "More" respectively. For the "Less" case, note that this is equivalent to $j \ge i + 2$. We therefore need there to be at least two $b$s. Then, we want a number of $a$s, which are simultaneously matched by additional $b$s, and finally, we can finish by adding as many extra $b$s as we want. The following rules capture this:
\begin{align*}
L &\to L'bb & \text{Add the mandatory two $b$s} \\
L' &\to aL'b & \text{Add as many $a$s as extra $b$s} \\
L' &\to T & \text{Once you've finished adding $a$s, move on}\\
T &\to Tb & \text{Add more $b$s as desired, but no more $a$s} \\
T &\to \varepsilon & \text{Terminate}
\end{align*}
Similarly, for the "More" case, this boils down to $i \ge j$. We get a similar setup:
\begin{align*}
M &\to aMb & \text{Add as many $a$s as $b$s} \\
M &\to V & \text{Once you've finished adding $a$s and $b$s, move on}\\
V &\to aV & \text{Add more $a$s as desired, but no more $b$s} \\
V &\to \varepsilon & \text{Terminate}
\end{align*}
In total, we can even optimise it a bit by removing the $L'$ state. See if you can see how to do that.
For the second one, we can start by adding as many $a$s as we want, so long as we match it with $c$s. Then we can add as many $b$s as we want, again so long as we match it with $c$s. The grammar for this looks like this:
\begin{align*}
S &\to aSc & \text{Add in $a$s matched by the same number of $c$s} \\
S &\to T & \text{Move on when ready} \\
T &\to bTc & \text{Add in $b$s matched by the same number of $c$s}\\
T &\to \varepsilon & \text{Terminate}
\end{align*}
Suppose there was a string $s \in (a | b)^*$ such that $s \notin L_1$ and $s \neq a^nb^n$ for any natural $n,$ where $s$ is of the smallest size such that there is a string which meets these conditions. ($s$ is minimal)
Clearly $|s|$ must be $1, 2,$ or greater than or equal to $2.$ The case where $|s| = 1$ is covered trivially: $a$ and $b$ are in $L_1.$
For $|s| = 2, L_1$ generates $aa$ via $S \to Sa \to aa, ba$ via $S \to bS \to ba,$ and $bb$ via $S \to bS \to bb.$ The only remaining $s$ such that $|s| = 2$ is $ab,$ but this would clearly contradict the definition of $s$ since $ab = a^1b^1.$ So, we must have $|s| > 2.$
Consider the decomposition $s = c_1wc_2,$ for $c_1, c_2 \in \{a,b\}, w \in (a|b)^*.$ Because our grammar has the rules $S \to \{aSa, aSb, bSa, bSb\},$ if $w$ is generated by our grammar then $s$ must be as well. So, because $s$ is not generated by our grammar, we must have that $w$ is not generated by our grammar, and because $s$ is minimal, we must have that $w$ is of the form $a^n b^n,$ otherwise we would have a smaller counterexample. This gives us three forms to consider for $s:$ $s = a(a^nb^n)a, s = b(a^nb^n)a,$ or $s = b(a^nb^n)b,$ noting that the form $a(a^nb^n)b = a^{n+1}b^{n+1}$ contradicts our definition of $s.$
Now, consider a string $x \in (a|b)^*$ with an odd length. By our argumentation before, if we can generate such an $x$ with length $n,$ we can generate any $x$ with length $n+2,$ so because we can generate any such $x$ with $|x| = 1,$ by induction we can show that we can generate any such $x$ with $|x| = 2k + 1, k > 0.$
So, going through our forms for $s,$ rewrite $a(a^nb^n)a$ as $(aa^nb^n)a = xa$ with $|x| = 2n+1$ for some natural $n.$ Because of our rule $S \to Sa,$ clearly we can generate strings of this form. Similarly, we can generate strings of the forms $b(a^nb^n)a = (ba^nb^n)a = xa$ and $b(a^nb^n)b = b(a^nb^nb) = bx.$ We have now covered every possible form for $s,$ showing each one either can be generated or is of the form $a^nb^n,$ so we have our contradiction and there can be no such minimal counterexample $s.$
Hope this helps!
Best Answer
Have a non-terminal symbol $A$ that generates any string of two terminal symbols and disappears. If $S$ is the initial symbol, have a production $S\to AB$, where $B\to aBa\mid bBb\mid aAa\mid bAb$.