An investment of N2800 earns N128 per annum. Part of the investment earns 5% and the rest earns 4%. How much is invested at 5%?

Let $x$ be the amount invested at 5% and $y$ be the amount invested at 4%.
The total sum invested is N2800, so:

$$x + y = 2800 \quad (1)$$

The total income is N128 per annum. The income from $x$ is $0.05x$ and from $y$ is $0.04y$.

$$0.05x + 0.04y = 128 \quad (2)$$

From equation (1), we can express $y$ in terms of $x$:

$$y = 2800 - x$$

Substitute $y = 2800 - x$ into $0.05x + 0.04y = 128$:

$$0.05x + 0.04(2800 - x) = 128$$ $$0.05x + (0.04 \times 2800) - 0.04x = 128$$ $$0.05x + 112 - 0.04x = 128$$

Combine the terms with $x$:

$$(0.05 - 0.04)x + 112 = 128$$ $$0.01x + 112 = 128$$

Subtract 112 from both sides:

$$0.01x = 128 - 112$$ $$0.01x = 16$$

Divide by 0.01:

$$x = \frac{16}{0.01}$$ $$x = 1600$$

The amount invested at 5% is N1600.

The final answer is $\text{N1600}$.