Answer
(a) The tension in the rope holding the pulley is $\sqrt{2}~Mg$
(b) $\theta = 45^{\circ}$
Work Step by Step
(a) Let $T_w$ be the tension in the rope holding the weight. Then the vertical section of this rope has a tension of $T_w = Mg$ and the horizontal section of this rope has a tension of $T_w = Mg$.
Let $T_p$ be the tension in the rope holding the pulley. Then the horizontal component of $T_p$ is equal in magnitude to $T_w$ and the vertical component of $T_p$ is equal in magnitude to $T_w$.
We can find $T_p$:
$T_p = \sqrt{T_w^2+T_w^2} = \sqrt{2}~T_w = \sqrt{2}~Mg$
(b) We can find the angle $\theta$ that the rope makes with the ceiling:
$tan~\theta = \frac{T_{p,y}}{T_{p,x}} = \frac{T_w}{T_w}$
$tan~\theta = 1$
$\theta = tan^{-1}~1$
$\theta = 45^{\circ}$
