Identify forces and set up equilibrium equations.

Let $R_A$ be the normal reaction force at the floor (point A).
Let $F_A$ be the friction force at the floor (point A).
Let $R_B$ be the normal reaction force at the wall (point B).
The weight of the ladder $W$ acts at its center, $a$ from A, as it is uniform and has length $2a$.
The ladder makes an angle of $45^\circ$ with the horizontal.

For vertical equilibrium (sum of vertical forces = 0):

For horizontal equilibrium (sum of horizontal forces = 0):

Taking moments about point A eliminates $R_A$ and $F_A$ from the equation.
The weight $W$ creates a clockwise moment. The perpendicular distance from A to the line of action of $W$ is $a \cos 45^\circ$.
The normal force $R_B$ creates an anticlockwise moment. The perpendicular distance from A to the line of action of $R_B$ is $2a \sin 45^\circ$.
Sum of moments about A = 0:

Since $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$:

From horizontal equilibrium, $F_A = R_B = \frac{W}{2}$.
For limiting equilibrium, the friction force is $F_A = \mu R_A$.
Substitute the values of $F_A$ and $R_A$:

The value of $\mu$ is $\boxed{\frac{1}{2}}$.

Let the boy of weight $2W$ be at a distance $x$ from A along the ladder.
The ladder is on the verge of slipping, so the friction force $F_A$ is at its maximum value, $F_A = \mu R_A$. We use $\mu = \frac{1}{2}$ from part (a).

For vertical equilibrium:

For horizontal equilibrium:

Since the ladder is about to slip, $F_A = \mu R_A$.

From horizontal equilibrium, $R_B = F_A = \frac{3W}{2}$.

The weight of the ladder $W$ creates a clockwise moment. Perpendicular distance is $a \cos 45^\circ$.
The weight of the boy $2W$ creates a clockwise moment. Perpendicular distance is $x \cos 45^\circ$.
The normal force $R_B$ creates an anticlockwise moment. Perpendicular distance is $2a \sin 45^\circ$.
Sum of moments about A = 0:

Since $\sin 45^\circ = \cos 45^\circ$:

Substitute $R_B = \frac{3W}{2}$:

The boy can climb a distance of $\boxed{a}$ from end A.