If $M$ is a homotopy sphere of dimension $4k>0$, then the signature is clearly zero. By the Hirzebruch signature theorem, you get $0=\langle L_k (TM); [M] \rangle = b_k \langle p_k (TM); [M] \rangle$ for a certain number $b_k \neq 0$. Therefore, the Pontrjagin classes of $TM$ are all trivial, and hence the $\hat{A}$-genus is zero. This settles part (1).

Part (2) is harder. Adams proved in his $J(X)$ papers (it is one of the main results stated in the introduction to part IV) that the unit map from the sphere spectrum to $KO$ is surjective in homotopy of degrees $8k+1$ and $8k+2$. The unit map $\mathbb{S} \to KO$ factors as

$$\mathbb{S} \to MSpin \to KO$$

with the first map the unit and the second the Atiyah-Bott-Shapiro orientation. All that is needed to show this claim is that the $\alpha$-invariant is a point (this is a statement about the homomorphism $\pi_0 (MSpin) \to \pi_0 (KO)$).

Therefore, there exists framed manifolds of dimension $8k+1$ and $8k+2$ whose $\alpha$-invariants (as spin manifolds) are zero.

Now use the work of Kervaire-Milnor. They prove in ''Groups of homotopy spheres'' that each odd-dimensional framed manifold is framed cobordant to a homotopy sphere, if the dimension is at least $5$. Take a framed manifold of dimension $8k+1$ with nonzero $\alpha$ and replace it by a homotopy sphere cobordant to it.
Therefore, $\beta_n$ is surjective if $n = 8k+1$ (for $k=0$, this is also true). 

In dimensions $8k+2$, a framed manifold is framded cobordant to a homotopy sphere if and only if its Kervaire invariant is zero. This is why I do not see a short argument in that case.