For the first question, yes. It is consistent to have that.

- Solovay's model, or any model of $\sf AD$ for example.
- Truss' models, which are similar to Solovay's model, only we start with a general limit cardinal (rather than an inaccessible). The result has that $\aleph_1$ is singular, but still cannot be embedded into $\Bbb R$.
- The Feferman-Levy model where $\Bbb R$ is a countable union of countable sets. This again has a striking similarity to the previous two models with this and the Truss model generated by collapsing $\aleph_\omega$ being very different despite having very similar constructions.

For your second question it is also easily achievable for every $\lambda$ whose cofinality is not $\omega$. Simply start with a model where $2^{\aleph_0}=\lambda$, then add any infinite number of Cohen reals and make them a non-well orderable set. This is a simple variation of Cohen's first model which one can find in many places in the literature.

Since the Cohen forcing doesn't collapse cardinals, it is easy to show that any subset of the real numbers in the full generic extension has size $\leq\lambda$, and that in the symmetric extension itself not all the sets of reals can be well-ordered. The ground model reals witness a set of size $\lambda$.

Finally, the third question is an interesting one. But the answer is no. This is a nice application of the "Division by three" paper by Conway and Doyle. There they show that without the axiom of choice it holds that if $|A\times 3|=|B\times 3|$ then $|A|=|B|$. They remark that the claim is also true when replacing $3$ by $2$. Now we will have: $$|A|+|A|=|\Bbb R|=|\Bbb R|+|\Bbb R|\implies |A|=|\Bbb R|.$$

I will add the remark, though, that $\Bbb R$ can consistently be partitioned into two sets of strictly smaller cardinality. But they will not be equipotent.