a.T=$\{0^n1^m0^n|m,n\geq 0\}$
Proof Let P be the pumping length, so $\exists$ a string w such that $|w|\geq p$. Let $w=0^p1^t0^p$ for $t \geq 0$ and $2p+t>p$. Let $xy=0^p$ and $|xy|\leq p$, so $x=0^{p-1}$ and $y=0^1$. We can see that by the pumping lemma $xy^2z \notin T$ because there is an extra zero in the left side which leads to a contradiction.
b. $T=\{0^m1^n|m \neq n\}$ $T^{'}=\{0^m1^n|m=n \}$
Let p be the pumping length of $T^{'}$, so $\exists$ a string w such that $|w| \geq p$. Let $w=0^p1^p$ since $2p >p$. Let $xy=0^p$ since $|xy|\leq p$, so $x=0^{p-1}$ and $y=0$. We can see that $xy^2z \notin T^{'}$ because $0^{p-1}0^2 1^p=0^{p+1}1^p$, so there is more 0's than 1's. This leads to a contradiction, so $T^{'}$ is not regular which implies T is not regular
c. $T=\{w|w \in \{0,1\}^{*} \text{is not a palindrome} \}$ $T^{'}=\{w|w \in \{0,1\}^{*} \text{is a palindrome} \}$
Let $p$ be the pumping length of $T^{'}$, so $\exists$ a string w such that $|w| \geq p$. Let $w=0^{p}1^{p}0^{p}$. It's easy to see that w is a palyndrome, so let $xy=0^p$. This works because $|xy| \leq p$, so we can set $x=0^{p-1}$ and $y=0$. It's evident that $xy^2z=0^{p-1}0 1^p0^p =0^{p+1}1^p0^p \notin T^{'}$ because $0^{p+1}1^p 0^p \neq 0^p1^p0^{p+1}$, so contradiction. Therefore, $T^{'}$ is regular which implies T is regular.
d. $T=\{wtw| w,t \in \{0,1\}^{+}$
If $w \in \{0,1\}^{+}$ then $w^n \in \{0,1\}^{+}$ for $n \geq 1$
If $t \in \{0,1\}^{+}$ then $t^n \in \{0,1\}^{+}$ for $n \geq 1$
Let p be the pumping length of $T$, so $\exists \phi$ such that $|\phi| \geq p$. Let $\phi =w^pt^pw^p$ since $|\phi| \geq p$, so let $xy=w^p$ and $z=t^pw^p$. This works because $|xy| \leq p$. It's evident that $xy^2z=w^pwt^pw^p=w^{p+1}t^{p}w^{p} \not in T$ because $w^{p+1} \neq w^p$, so this contradicts the pumping lemma. Therefore; T is not regular
How does that look?