Collision resistant hash function A function is $(\varepsilon, t)$-collision resistant if there is no boolean circuit (using "not", "and", "or") of size at most $t$ which outputs a collision with probability at least $\varepsilon$.

Let $h_0:\{0,1\}^{2m}\rightarrow\{0,1\}^m$ be a $(\varepsilon, t)$-collision resistant hash function and $i\in\mathbb{N}_{\geq 1}$.
Define a hash function $h_i: \{0,1\}^{2^{i+1}\cdot m}\rightarrow \{0,1\}^m$ recursively using $h_{i-1}$ in the following way:
Interpret the bit string $x\in \{0,1\}^{2^{i+1}\cdot m}$ as $x=x_1 x_2$, where both $x_1,x_2\in \{0,1\}^{2^i\cdot m}$.  

Then the hash value $h_i(x)$ is defined as

$$h_i(x)=h_0(h_{i-1}(x_1)h_{i-1}(x_2)).$$

For which $(\varepsilon_i, t_i)$ is $h_i$ $(\varepsilon_i, t_i)$-collision resistant?
And can we successfully use the birthday attack on this hash function to find collisions?
The birthday attack applies to any hash-function, regardless of its structure, and also to $h_i$.

For your first question, here is a possible hint. Note that


if we can find a collision in $h_i$ then we can find a collision in $h_0$ (by considering the top-most $h_0$, for instance).
if we can find a collision in $h_0$ then, we can find a collision for $h_i$ (by replacing one of the inner-most $h_0$ with it's collision).