A reactant R falls from 0.800 mol/L at t = 0 to 0.520 mol/L at t = 90 s…

Step 1. The average rate. $\dfrac{0.800\ \text{mol/L} - 0.520\ \text{mol/L}}{90\ \text{s}} = \dfrac{0.280\ \text{mol/L}}{90\ \text{s}} = 3.11 \times 10^{-3} = 0.00311\ \text{mol L}^{-1}\ \text{s}^{-1}$.

Step 2. The slope of the tangent. It falls by $0.744\ \text{mol/L}$ over $299\ \text{s}$. The rate is $\dfrac{0.744\ \text{mol/L}}{299\ \text{s}} = 2.49 \times 10^{-3} = 0.00249\ \text{mol L}^{-1}\ \text{s}^{-1}$.

Step 3. Why they differ. The average is taken over a whole interval. The reaction was faster at the start of the interval, when the concentration was higher, than at $90\ \text{s}$. So the average is larger than the rate at the end.