The pumping lemma for CFL states: If $L$ is context-free, then there exists a positive integer $P> 0$, such that for any $x\in L$ with $\left | x \right |\geq P$, there exists $v,w,x,y,z$ such that $u=vwxyz$, $\left | wxy \right |\leq P$, $\left | wy \right |> 0$ and $\forall m\geq 0$, $uw^{m}xy^{m}z\in L$.
$a)$ Suppose for $L=\left \{ 0^{n}1^{n}0^{n}1^{n} \right \}$ $\left | wy \right |$ non-empty where $u=\epsilon$ and $z=\epsilon$ , $L=\left \{{\color{Red}{0^{n}} } 1^{n}0^{n}{\color{Red} {1^{n}}} \right \}$
then $\left | x \right |=1^{n}0^{n}$
But according to 3rd pumping lemma condition $\left | wxy \right |\leq n$
But it is not in the form $uw^{m}xy^{m}z\in L$, where loop exists for $w$ and for $y$ That is not possible for this language. So, it is non CFL.
$b)$ If we divide $L=uvxyz$ in $3$ segment, then atleast one segment doesnot in $v$ or in $y.$ Hence $xv^{2}xy^{2}z$ doesnot maintain a ratio of $1:2:3$. So, they cannot be pumped by pumping lemma.
$c)$ Let $L=a^{p}b^{p}$#$a^{p}b^{p}$
We show that string $L=uvxyz$ cannot be pumped using pumping lemma.
Say neither $v$ , nor $y$ contains #, otherwise $uv^{0}xy^{0}z$ not in $L$
If $v$ and $y$ are non-empty and occur in left hand side of #, then $uv^{2}xy^{2}z$ cannot be in $L.$ Because it is longer than left hand side of #.
Similarly , if $v$ and $y$ are non-empty and occur in right hand side of #, then $uv^{2}xy^{2}z$ cannot be in $L.$ Because it is longer than right hand side of #
Only, remaining case is when $v$ and $y$ are non-empty and straddle the #. but because the 3rd pumping lemma condition $\left | vxy \right |\leq p$
Hence $uv^{2}xy^{2}z$ contains more $0's$ on right hand side of #. So, it not in $L.$
$d)$ Now we can take $t_{1}=a^{p}b^{p}$,$t_{2}=a^{p}b^{p}$
So, $L=a^{p}b^{p}$#$a^{p}b^{p}$
Now proving is same as above.
