Given vectors OA = a and OB = b, with Q being the midpoint of OA and P on AB such that AP = 3PB. Express AB, BQ, OP in terms of a and b.

Here are the solutions to the vector problem:

a) Express in terms of $\mathbf{a}$ and $\mathbf{b}$:

i. $\vec{AB}$

$$\vec{AB} = \vec{OB} - \vec{OA}$$ $$\vec{AB} = \mathbf{b} - \mathbf{a}$$

The vector $\vec{AB}$ is $\boxed{\mathbf{b} - \mathbf{a}}$.

ii. $\vec{BQ}$
Given that Q is the midpoint of OA,

$$\vec{OQ} = \frac{1}{2}\vec{OA} = \frac{1}{2}\mathbf{a}$$

Now, find $\vec{BQ}$:

$$\vec{BQ} = \vec{OQ} - \vec{OB}$$ $$\vec{BQ} = \frac{1}{2}\mathbf{a} - \mathbf{b}$$

The vector $\vec{BQ}$ is $\boxed{\frac{1}{2}\mathbf{a} - \mathbf{b}}$.

iii. $\vec{OP}$
Given that P is a point on AB such that $\vec{AP} = 3\vec{PB}$. This means P divides AB in the ratio 3:1.
So, $\vec{AP} = \frac{3}{4}\vec{AB}$.

$$\vec{OP} = \vec{OA} + \vec{AP}$$ $$\vec{OP} = \vec{OA} + \frac{3}{4}\vec{AB}$$

Substitute $\vec{AB} = \mathbf{b} - \mathbf{a}$ from part (a)i:

$$\vec{OP} = \mathbf{a} + \frac{3}{4}(\mathbf{b} - \mathbf{a})$$ $$\vec{OP} = \mathbf{a} + \frac{3}{4}\mathbf{b} - \frac{3}{4}\mathbf{a}$$ $$\vec{OP} = \left(1 - \frac{3}{4}\right)\mathbf{a} + \frac{3}{4}\mathbf{b}$$ $$\vec{OP} = \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}$$

The vector $\vec{OP}$ is $\boxed{\frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}}$.

b) Given further that $\vec{OX} = k\vec{OP}$ and $\vec{BX} = h\vec{BQ}$, give two expressions of $\vec{OX}$ in terms of $\mathbf{a}$, $\mathbf{b}$, $k$ and $h$.

First expression for $\vec{OX}$ (using $\vec{OX} = k\vec{OP}$):
Substitute the expression for $\vec{OP}$ from part (a)iii:

$$\vec{OX} = k\left(\frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\right)$$ $$\vec{OX} = \frac{k}{4}\mathbf{a} + \frac{3k}{4}\mathbf{b}$$

Second expression for $\vec{OX}$ (using $\vec{BX} = h\vec{BQ}$):
We know that $\vec{OX} = \vec{OB} + \vec{BX}$.
Substitute $\vec{BX} = h\vec{BQ}$ and the expression for $\vec{BQ}$ from part (a)ii:

$$\vec{OX} = \vec{OB} + h\vec{BQ}$$ $$\vec{OX} = \mathbf{b} + h\left(\frac{1}{2}\mathbf{a} - \mathbf{b}\right)$$ $$\vec{OX} = \mathbf{b} + \frac{h}{2}\mathbf{a} - h\mathbf{b}$$ $$\vec{OX} = \frac{h}{2}\mathbf{a} + (1 - h)\mathbf{b}$$

The two expressions for $\vec{OX}$ are:

1. $\boxed{\vec{OX} = \frac{k}{4}\mathbf{a} + \frac{3k}{4}\mathbf{b}}$
2. $\boxed{\vec{OX} = \frac{h}{2}\mathbf{a} + (1 - h)\mathbf{b}}$

c) Find the values of $h$ and $k$.

Equate the two expressions for $\vec{OX}$ from part (b):

$$\frac{k}{4}\mathbf{a} + \frac{3k}{4}\mathbf{b} = \frac{h}{2}\mathbf{a} + (1 - h)\mathbf{b}$$

Since $\mathbf{a}$ and $\mathbf{b}$ are non-parallel vectors, the coefficients of $\mathbf{a}$ and $\mathbf{b}$ on both sides must be equal.

Equating coefficients of $\mathbf{a}$:

$$\frac{k}{4} = \frac{h}{2}$$ $$k = \frac{4h}{2}$$ $$k = 2h \quad \text{(Equation 1)}$$

Equating coefficients of $\mathbf{b}$:

$$\frac{3k}{4} = 1 - h \quad \text{(Equation 2)}$$

Substitute Equation 1 into Equation 2:

$$\frac{3(2h)}{4} = 1 - h$$ $$\frac{6h}{4} = 1 - h$$ $$\frac{3h}{2} = 1 - h$$

Multiply both sides by 2:

$$3h = 2(1 - h)$$ $$3h = 2 - 2h$$ $$3h + 2h = 2$$ $$5h = 2$$ $$h = \frac{2}{5}$$

Now substitute the value of $h$ back into Equation 1 to find $k$:

$$k = 2h$$ $$k = 2\left(\frac{2}{5}\right)$$ $$k = \frac{4}{5}$$

The values are $\boxed{h = \frac{2}{5}, k = \frac{4}{5}}$.