between Borel $\sigma$ algebra and Lebesgue $\sigma$ algebra, are there any other $\sigma$ algebra?

First question (title): Sure. It is not hard to show that the sets of the form $B \cup S$, where $B$ is Borel and $S$ is a subset of the Cantor-set constitute a $\sigma$-algebra. There are $2^{\mathfrak{c}}$ subsets of the Cantor set but only $\mathfrak{c}$ Borel sets, hence this $\sigma$-algebra lies strictly between the Borel sets and the Lebesgue-measurable sets.

It's very rare that there is no $\sigma$-algebra strictly between two $\sigma$-algebras, it essentially means that there are atoms (sets that cannot be written non-trivially as union of proper subsets.

I'm not sure I understand the last question. You can take for instance Lebesgue measure plus a Dirac measure if you want something strictly different (i.e. not related via Radon-Nikodym).

Your first question has been answered by Theo Buehler; for your second question, you can always take $\mathcal{P}(\mathbb{R})$ as a $\sigma$-algebra; a nontrivial example is the $\sigma$-algebra of all subsets of $\mathbb{R}$ that are either countable or co-countable (that is, all $X$ for which either $|X|\leq\aleph_0$ or $|\mathbb{R}-X|\leq\aleph_0$; it is straightforward to verify this is a $\sigma$-algebra).

For your last question: if you place no restrictions whatsoever on the measure, then the answer is trivial: yes, there are other measures. For a trivial example, just scale the Lebesgue measure by a factor different from $1$.

More generally, given a measurable positive function $f$, define the measure $\mu$ on the Lebesgue measurable sets by $\mu(X) = \int_X f d\lambda = \int f\chi_X d\lambda$, where $\lambda$ is the Lebesgue measure and $\chi_X$ is the characteristic function of $X$. This is a measure:

1. If $X$ is any measurable set, then $\mu(X) = \int_X f\,d\lambda \geq 0$ because $f(x)\geq 0$ for all $x$, so $\mu$ is nonnegative.

2. If $\{E_i\}$ is a countable collection of pairwise disjoint sets, then $$\mu(\cup E_i) = \int_{\cup E_i}f\,d\lambda = \sum_{i=1}^{\infty} \int _{E_i} f\,d\lambda = \sum_{i=1}^{\infty}\mu(E_i)$$ (with measures and sum possibly infinite); and

3. $\mu(\emptyset) = \int_{\emptyset}f\,d\lambda = 0$.

Thus, $\mu$ is a measure on the Lebesgue $\sigma$-algebra; unless $f(x)=1$ for almost all $x$, you have $\mu\neq\lambda$. It may even be a finite measure, if $f\in\mathcal{L}^1$.