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(a) Differentiating both sides of

$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$

gives

$$\sum_{n=0}^\infty n x^{n-1}=\frac{1}{(1-x)^2}$$

Now multiplying both sides by $x$ gives

$$\sum_{n=0}^\infty n x^n=\frac{x}{(1-x)^2}$$

as desired.

(b) Differentiating both sides of the conclusion of part (a) gives

$$\sum_{n=0}^\infty n^2 x^{n-1}=\frac{(1-x)^2-2x(x-1)}{(1-x)^4} =\frac{(1-x)(1-x+2x)}{(1-x)^4} =\frac{1+x}{(1-x)^3}$$

Now multiplying both sides by $x$ gives

$$\sum_{n=0}^\infty n^2 x^n=\frac{x(1+x)}{(1-x)^3}$$

We know that differentiation preserves the radius of convergence of power series. So this series has radius of convergence $1$ (the radius of convergence of the original geometric series). At $x=\pm1$ the series diverges by the divergence test. So the series converges for $-1<x<1$.