(a). Suppose we need to reach the $n^{th}$ step , we can reach so from the $n-1^{th}$ step or from the $n-2^{th}$ step.

Let $a_n$ be the number of ways we can reach the $n^{th}$ step.

Therefore $a_n=a_{n-1}+a_{n-2}$ which follows from my initial statement.

(b) $a_0=0,a_1=1$. Trivially .

(c). The characteristic equation of the recurrence relation is of the form : $r^2-r-1=0$. The roots are $1+\frac{\sqrt{5}}{2}$,$1-\frac{\sqrt{5}}{2}$. Therefore closed form solution of the recurrence relation is : $a_n$=$\alpha_{0}(1+\frac{\sqrt{5}}{2})^{n}$ + $\alpha_{1}(1-\frac{\sqrt{5}}{2})^{n}$

$\because$ $a_0$=0. We have 0=$\alpha_0+\alpha_1$ $\implies$ $\alpha_0=-\alpha_1$.

$a_1$=$\alpha_{0}(1+\frac{\sqrt{5}}{2})$ + $\alpha_{1}(1-\frac{\sqrt{5}}{2})$ $\implies$$a_1$=$-\alpha_{1}(1+\frac{\sqrt{5}}{2})$ + $\alpha_{1}(1-\frac{\sqrt{5}}{2})$ $\implies$ $a_1=-2 \times \frac{\sqrt{5}}{2}\alpha_1$=$-\sqrt{5}\alpha_1$

$\because a_1=1 \therefore \alpha_1$=$-\frac{1}{\sqrt{5}} \implies \alpha_0= \frac{1}{\sqrt{5}}$

$\therefore a_n=\frac{1}{\sqrt{5}}(1+\frac{\sqrt{5}}{2})^n-\frac{1}{\sqrt{5}}(1-\frac{\sqrt{5}}{2})^n$