Let $\lambda=\frac{1+\sqrt{5}}{2}$ and $\mu=\frac{1-\sqrt{5}}{2}$. You are right in asserting that there are constants $a$ and $b$ such that $L_n=a\lambda^n +b\mu^n.\tag{$1$}$
To find $a$ and $b$, use the initial conditions. But because we are lazy, we will use a little trick. If we extend the sequence backwards, we see that if the recurrence is to be satisfied, $L_0$ must be $2$. The formula, by general theory, should hold at $n=0$, so we get $a\lambda^0+b\mu^0=2$. It follows that $a+b=2$.
From the fact that $L_1=1$, we find, substituting in $(1)$, that $a\lambda+ b\mu =1$. This gives $a\left(\frac{1+\sqrt{5}}{2} \right)+b\left(\frac{1-\sqrt{5}}{2} \right)=1.$ Expand. Already we have $\frac{a}{2}+\frac{b}{2}=\frac{2}{2}=1$. So the part with the $\sqrt{5}$'s must be equal to $0$. That implies that $a-b=0$.
So $a+b=2$ and $a-b=0$, giving $a=b=1$. We conclude that
$L_n=\lambda^n +\mu^n.$
Remark: On the confusion about the $n\ge 3$ part: We are told $L_1$ and $L_2$ outright. Then there is the recurrence $L_n=L_{n-1}+L_{n-2}$, for $n \ge 3$. Put in particular $n=3$. That tells us that $L_3=L_2+L_1=4$. Now put $n=4$. That tells us that $L_4=L_3+L_2=7$. And so on forever. The $n \ge 3$ part is because there is no need to assume that the recurrence $L_n=L_{n-1}+L_{n-2}$ holds for $n \lt 3$, since we know $L_1$ and $L_2$.
Actually, the little trick we used to simplify the calculations finds $L_0$, on the assumption that the recurrence holds for $n=2$. And the Lucas sequence can be extended backwards, so that $L_n$ becomes defined for all integers. The formula for $L_n$ that we derived will then hold even for negative $n$.