Two exercise books and 3 pencils cost KSh 190. 5 similar exercise books and two similar pencils cost KSh 290. Find the cost of each exercise book and each pencil.

Let $p$ be the cost of one pencil and $e$ be the cost of one exercise book.
From the problem statement, we have:

1. Two exercise books and three pencils cost KSh 190:

$$3p + 2e = 190 \quad (1)$$

2. Five similar exercise books and two similar pencils cost KSh 290:

$$2p + 5e = 290 \quad (2)$$

The student has already started by expressing $e$ from equation (1):

$$2e = 190 - 3p$$ $$e = \frac{190 - 3p}{2}$$

Substitute $e = \frac{190 - 3p}{2}$ into $2p + 5e = 290$:

$$2p + 5\left(\frac{190 - 3p}{2}\right) = 290$$

Multiply the entire equation by 2 to eliminate the fraction:

$$2(2p) + 2 \cdot 5\left(\frac{190 - 3p}{2}\right) = 2(290)$$ $$4p + 5(190 - 3p) = 580$$ $$4p + 950 - 15p = 580$$ $$-11p + 950 = 580$$ $$-11p = 580 - 950$$ $$-11p = -370$$ $$p = \frac{-370}{-11}$$ $$p = \frac{370}{11}$$

The cost of one pencil is KSh $\frac{370}{11}$.

$$e = \frac{190 - 3p}{2}$$ $$e = \frac{190 - 3\left(\frac{370}{11}\right)}{2}$$ $$e = \frac{190 - \frac{1110}{11}}{2}$$

To simplify the numerator, find a common denominator:

$$e = \frac{\frac{190 \times 11}{11} - \frac{1110}{11}}{2}$$ $$e = \frac{\frac{2090}{11} - \frac{1110}{11}}{2}$$ $$e = \frac{\frac{2090 - 1110}{11}}{2}$$ $$e = \frac{\frac{980}{11}}{2}$$ $$e = \frac{980}{11 \times 2}$$ $$e = \frac{980}{22}$$ $$e = \frac{490}{11}$$

The cost of one exercise book is KSh $\frac{490}{11}$.

The cost of one pencil is $\boxed{\text{KSh } \frac{370}{11}}$ and the cost of one exercise book is $\boxed{\text{KSh } \frac{490}{11}}$.