$L=\{u\$w^R|u,w\in \{a,b\}^+ \text{$w$ is prefix and suffix of $u$ }\}$
Check if language $L$ and $L^C$ is context free.
L
$a^*b^*\$a^*b^*\cap L = \{a^ib^j\$a^ib^j|i, j\ge 0\}\notin CFG$
So, $L$ is not context free.
And now look at, $L^C$. I claim that I can give you a grammar :)
Cases:
$(1)\{u\$w^R|u,w\in \{a,b\}^+ \text{$w$ is not prefix of $u$ }\}$
$(2)\{u\$w^R|u,w\in \{a,b\}^+ \text{$w$ is not sufixf of $u$ }\}$
$(3)\{w|\text{ $w$ doeas not contain \$}\}$
$(4)\{w|\text{ $w$ contains two or more \$s}\}$
$(4): C\rightarrow X\$X\$X$
$\ \ \ \ \ \ \ \ X\rightarrow \epsilon|aX|bX$
$(3):T\rightarrow X$
$(1):S\rightarrow AXSXB|B'XSXA'|\$$
$A\rightarrow bA|aA|a $
$B\rightarrow Bb|Ba|b $
$A'\rightarrow Ab|Aa|a $
$B'\rightarrow bB|aB|b$
$(2)$ is analogical to $(1)$.
What do you think about it?
There is a $^R$ for the $w$ so the intersection should be with $a^*b^*\$b^*a^*$, shouldn't it?
For case 4, your grammar does not produce words with "more" than 2 $\$$.
For case 1, it seems ok. But an easier gramar is:
S -> aSa | bSb | aX\$Xb | bX\$Xa | \$.
Otherwise your solution seems all right.