$\displaystyle \frac{a^2-9}{a^2-3a}\cdot\frac{a^2+a}{a+3} \implies \frac{(a+3)(a-3)}{a(a-3)}\cdot\frac{a(a+1)}{a+3}\implies a+1$

For the second question, I will think outside the box:

Since $\displaystyle 4=2^2$, then

$\displaystyle 2^x+2=2^{2x}-1\implies 2^{2x}-2^x-3=0$

Making a substitution $\displaystyle u=2^x$, we get the quadratic:

$\displaystyle u^2-u-3=0 \implies u=\frac{1\pm\sqrt{1+12}}{2}$.

Thus,

$\displaystyle 2^x=\frac{1\pm\sqrt{1+12}}{2}$. We ignore the negative answer, because $\displaystyle 2^x$ can NEVER be negative.

Thus, $\displaystyle 2^x=\frac{1+\sqrt{13}}{2} \implies x=\frac{1+\sqrt{13}}{2\log 2}$

Hope this helped you out!!