Answer
(a) Since the particle begins at the origin $O$, so at $t=T$, it is at $O$.
(b) If ${\bf{\bar v}} = 0$, then ${\bf{r}}\left( T \right) = {\bf{r}}\left( 0 \right)$. However, it does not imply that the velocity is zero. Hence, it is not necessary that the particle's average speed equals to zero.
Work Step by Step
(a) The average velocity vector is given by
${\bf{\bar v}} = \frac{1}{T}\mathop \smallint \limits_0^T {\bf{r}}'\left( t \right){\rm{d}}t$
Since ${\bf{r}}'\left( t \right) = \frac{{d{\bf{r}}}}{{dt}}$, we can write
${\bf{\bar v}} = \frac{1}{T}\mathop \smallint \limits_0^T {\bf{r}}'\left( t \right){\rm{d}}t = \frac{1}{T}\mathop \smallint \limits_0^T {\rm{d}}{\bf{r}}\left( t \right) = \frac{1}{T}\left( {{\bf{r}}\left( T \right) - {\bf{r}}\left( 0 \right)} \right)$
If ${\bf{\bar v}} = 0$, then ${\bf{r}}\left( T \right) = {\bf{r}}\left( 0 \right)$. This implies that at $t=T$, the particle is located at the same location as ${\bf{r}}\left( 0 \right)$. Since the particle begins at the origin $O$, so at $t=T$, it is at $O$.
(b) As is shown in part (a), if ${\bf{\bar v}} = 0$, then ${\bf{r}}\left( T \right) = {\bf{r}}\left( 0 \right)$. However, it does not imply that the velocity is zero. Hence, it is not necessary that the particle's average speed equals to zero.
