A) Your answer is not clear about whether linear momentum is conserved. As you state, there is an external force acting during the collision between the ball $m$ and the block $M$. This external force is the reaction at the hinge (pivot), which is attached to the outside world. This means that linear momentum might not be conserved. However, angular momentum about the hinge is conserved.
B) The question states that "the system" is at rest after the collision. Doesn't this mean the ball + rod? You have calculated the angular velocity immediately before the collision.
I can see that you have equated PE lost to KE gained, using $\frac12 I \omega^2$ for KE. But how have you calculated moment of inertia $I$?
The moment of inertia of the rod about one end is $\frac13 mL^2$, not $m(\frac12 L)^2=\frac14 mL^2$. The moment of the (point) mass $m$ is $mL^2$. So the total moment of inertia of ball+rod is $\frac43 mL^2$ not $\frac98 mL^2$.
Your calculation method for moment of inertia is wrong. The moment inertia of an extended object is not the moment of inertia of its CM. For example, if the rod were pivoted at its CM, your method gives MI=0 instead of $\frac{1}{12}mL^2$. However, the method does work for calculating the change in PE (provided the gravitational field is uniform).
C) Here you need to apply conservation of angular momentum not linear momentum.
The KE of the ball+rod immediately before the collision equals the KE of the block immediately after the collision :
$\frac12 I\omega^2=\frac12 Mu^2$
Also, the angular momentum of the ball+rod equals that of the block :
$I\omega=MuL$
so
$I^2\omega^2=M^2u^2L^2$
Dividing the AM equation by the KE equation we get
$I=ML^2$
However, $I=\frac43 mL^2$ therefore $M=\frac43 m$.
D) The velocity $u$ of the block immediately after the collision is given by $\frac12 Mu^2=32mgL$ where $M=\frac43m$ therefore $u=\frac32\sqrt{gL}$.
E) Your calculation of the velocity of the block as it leaves the table does not seem to be correct.
The initial KE of the block before it slides across the rough part of the table is $\frac12 Mu^2=\frac12 M \frac94 gL=\frac98 MgL$ from above. The work done on the block by friction is $\mu Mgd$. So the velocity $v$ of the block as it leaves the table is given by
$\frac12 Mv^2=\frac98 MgL-\mu Mgd$
$v=\sqrt{(\frac94L-2\mu d)g}$.
It is not possible to simplify this expression unless you are given a value for $\mu d$ in terms of $L$. EG if $\mu d=\frac58 L$ then $v=\sqrt{gL}$.
Judging by the diagram $D$ is measured from the edge of the table, not from the initial position of the block. Then $D=vt$ where $h=\frac12 gt^2$ which you have got right.
