Last digit of $3^{459}$.

Try out the first few powers of $3$: we have that $$ 3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243, 3^6=729,\ldots $$ It seems like the final digit cycles in a pattern, namely $$ 3\to9\to7\to1\to3\to9\to7\to1, $$ of length $4$. Since $$ 459=4\cdot114+3, $$ the final digit is the third in the cycle, namely $\color{red}{\mathbf{7}}$, and, sure enough, your Wolfram|Alpha computation confirms this.

$$3^{4}=1\pmod{10}\implies (3^{4})^{114}=1 \pmod{10}$$ So $$3^{459}=3^{4 \times 114}\cdot 3^{3}=3^{3}\pmod{10}=7\pmod {10}$$