Not all non-regular languages fail the test of the pumping lemma. Wikipedia has an annoyingly complex example of a non-regular language which can be pumped. So even if a language is non-regular, we may not be able to prove this fact using the pumping lemma.
But it turns out we can use the pumping lemma to prove your first language is not regular. I'm not sure about the second.
Claim: $L_1$ is non-regular.
Proof: By the pumping lemma. Let $p$ be the pumping length. (I'm going to use the alphabet $\{a,b\}$ rather than $\{0,1\}$.) If $p = 1$, then take the string $abbaa$, which is in $L_1$ and pump it to $aabbaa$ which is not in $L_1$, so $L_1$ would not be regular.
If $p > 1$, then take the string $a^pbba^{N}$. (We'll figure out what we want $N$ to be later.) Then consider any division of the string into $xyz$ where $x=a^{p-k}$, $y=a^k$, and $z=bba^{N}$.
Now let's pump this string $i$ times. (We'll figure out what we want $i$ to be later.) We get the string $xy^iz$, which gives $a^{p-k}a^{ik}bba^{N} = a^{p-k+ik}bba^{N}$.
Now let's back up. First, we picked $N$. Then, some choice of $k$ was made. Then, we picked $i$. We want to figure out what $N$ to pick so that, for any choice of $k \in [1,p]$, we can choose an $i$ that makes this string a palindrome by making the number of $a$s on the left equal the number on the right. (It will always have even length.)
So we want to always get that $p-k+ik=N$. If we play around with the math, we find that we should pick $N=p+p!$ and pick $i=p!/k+1$.
So to recap, we picked $N=p+p!$ and chose the string $a^pbba^N$. Then some choice of $k$ was made so that the string was made up of $a^{p-k}ybba^N$ where $y=a^k$. Then we picked $i=p!/k + 1$. We pumped the string to get $a^{p-k}y^{i}bba^N = a^{p-k}a^{ik}bba^N = a^{p-k+ik}bba^N$.
But we know that $p-k+ik = p-k+(\frac{p!}{k}+1)k = p-k+p!+k=p+p!$. And $N=p+p!$. So the number of $a$s on both ends is the same, so the string is an even-length palindrome, so it's not in $L_1$, so $L_1$ is not regular. $\square$
