Note that for fixed $t_0$, we have by the martingale representation theorem that

$$Y_{t_0} = \int_0^{t_0} I\bigl( s \in [0,t_0-1) \bigr) \, dB_s.$$

In particular that for $t_0\leq 1$ the indicator yields $0$ trivially. 
However, if we want to consider the process $(Y_t)$, then the integrand $I\bigl( s \in [0,t-1) \bigr)$ is not progressively measurable in the Brownian filtration (i.e. not measurable w.r.t. the $\sigma$-field $\mathcal{B}([0,s]) \otimes \mathcal{F}_s$). Thus it is not an Ito process and also not a semi-martingale as the martingale representation is unique. This answers [Q1] and [Q2].

As for [Q3], one classical example for a process with finite quadratic variation that is not a semimartingale is [fractional Brownian motion][1] with Hurst-parameter $h<1/2$. (It has even quadratic variation constant $0$).


  [1]: http://en.wikipedia.org/wiki/Fractional_Brownian_motion