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To solve this problem, we will use vector algebra. We are given a parallelogram \(OABC\) with the following information:

1. \( \overrightarrow{OA} = \mathbf{a} \) and \( \overrightarrow{OC} = \mathbf{c} \).
2. Point \( M \) lies on \( AB \) such that the ratio \( AM : MB = 1 : 3 \).
3. Point \( N \) lies on \( BC \) such that the ratio \( BN : NC = 1 : 2 \).
4. \( \overrightarrow{OX} = \lambda \overrightarrow{OB} \).
5. \( \overrightarrow{MX} = \mu \overrightarrow{MN} \).

We need to find the values of \( \lambda \) and \( \mu \).

### Step 1: Expressing vectors \( \overrightarrow{M} \) and \( \overrightarrow{N} \)

- Since \( M \) divides \( AB \) in the ratio \( 1 : 3 \), we can express the position vector of \( M \) as:
  \[
  \overrightarrow{OM} = \frac{1}{4} \overrightarrow{A} + \frac{3}{4} \overrightarrow{B}
  \]
  Since \( \overrightarrow{A} = \mathbf{a} \) and \( \overrightarrow{B} = \mathbf{a} + \mathbf{c} \):
  \[
  \overrightarrow{OM} = \frac{1}{4} \mathbf{a} + \frac{3}{4} (\mathbf{a} + \mathbf{c}) = \mathbf{a} + \frac{3}{4} \mathbf{c}
  \]

- Since \( N \) divides \( BC \) in the ratio \( 1 : 2 \), we can express the position vector of \( N \) as:
  \[
  \overrightarrow{ON} = \frac{2}{3} \overrightarrow{B} + \frac{1}{3} \overrightarrow{C}
  \]
  Since \( \overrightarrow{B} = \mathbf{a} + \mathbf{c} \) and \( \overrightarrow{C} = \mathbf{c} \):
  \[
  \overrightarrow{ON} = \frac{2}{3} (\mathbf{a} + \mathbf{c}) + \frac{1}{3} \mathbf{c} = \frac{2}{3} \mathbf{a} + \mathbf{c}
  \]

### Step 2: Expressing \( \overrightarrow{X} \) using \( \lambda \)

Given \( \overrightarrow{OX} = \lambda \overrightarrow{OB} \) and \( \overrightarrow{OB} = \mathbf{a} + \mathbf{c} \):
\[
\overrightarrow{OX} = \lambda (\mathbf{a} + \mathbf{c})
\]

### Step 3: Expressing \( \overrightarrow{MX} \) and \( \overrightarrow{MN} \)

- \( \overrightarrow{MX} = \overrightarrow{OX} - \overrightarrow{OM} \):
  \[
  \overrightarrow{MX} = \lambda (\mathbf{a} + \mathbf{c}) - \left(\mathbf{a} + \frac{3}{4} \mathbf{c}\right)
  \]
  \[
  \overrightarrow{MX} = (\lambda - 1) \mathbf{a} + \left(\lambda - \frac{3}{4}\right) \mathbf{c}
  \]

- \( \overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} \):
  \[
  \overrightarrow{MN} = \left(\frac{2}{3} \mathbf{a} + \mathbf{c}\right) - \left(\mathbf{a} + \frac{3}{4} \mathbf{c}\right)
  \]
  \[
  \overrightarrow{MN} = \left(\frac{2}{3} - 1\right) \mathbf{a} + \left(1 - \frac{3}{4}\right) \mathbf{c}
  \]
  \[
  \overrightarrow{MN} = -\frac{1}{3} \mathbf{a} + \frac{1}{4} \mathbf{c}
  \]

### Step 4: Setting up the equation using \( \mu \)

Given \( \overrightarrow{MX} = \mu \overrightarrow{MN} \), we equate:
\[
(\lambda - 1) \mathbf{a} + \left(\lambda - \frac{3}{4}\right) \mathbf{c} = \mu \left(-\frac{1}{3} \mathbf{a} + \frac{1}{4} \mathbf{c}\right)
\]

From this, we can equate coefficients of \( \mathbf{a} \) and \( \mathbf{c} \) separately:

1. Coefficient of \( \mathbf{a} \):
   \[
   \lambda - 1 = -\frac{\mu}{3}
   \]

2. Coefficient of \( \mathbf{c} \):
   \[
   \lambda - \frac{3}{4} = \frac{\mu}{4}
   \]

### Step 5: Solving the system of equations

From equation (1):
\[
\mu = -3(\lambda - 1) \implies \mu = -3\lambda + 3
\]

Substituting \( \mu \) in equation (2):
\[
\lambda - \frac{3}{4} = \frac{-3\lambda + 3}{4}
\]
\[
4\lambda - 3 = -3\lambda + 3
\]
\[
7\lambda = 6
\]
\[
\lambda = \frac{6}{7}
\]

Substitute \( \lambda \) back to find \( \mu \):
\[
\mu = -3\left(\frac{6}{7} - 1\right) = -3\left(\frac{-1}{7}\right) = \frac{3}{7}
\]

### Conclusion

The values of \( \lambda \) and \( \mu \) are:
\[
\lambda = \frac{6}{7}, \quad \mu = \frac{3}{7}
\]

Would you like further details or have any questions?

Here are some related questions you might find interesting:

1. How do we determine the point of division of a line segment in a given ratio using vectors?
2. What are the properties of a parallelogram that we can use to solve vector problems?
3. How can the section formula be applied to more complex geometric shapes using vectors?
4. What are some real-world applications of using vectors in solving geometric problems?
5. How does the concept of linear combinations of vectors help in finding points on a line?

**Tip:** When dealing with vector problems, always break down the vectors into known components and use the properties of geometric figures to set up equations.

Learn how to solve a vector algebra problem involving position vectors and ratios. This detailed solution uses vector algebra concepts and formulas to find unknown values. Suitable for high school students interested in advanced geometry and vector algebra.

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