Doing it from first principles, you have
$\mathrm{Var}(X)=\mathrm{E}\left[\left(X-\mathrm{E}(X)\right)^2\right]=\int_{-\infty}^\infty(x-\mu)^2f_X(x)~dx\;,$ where $\mu=\int_{-\infty}^\infty xf_X(x)~dx\;.$
The function $f_X(x)$ is even (i.e., symmetric about the $y$-axis), so
$\begin{align*} \int_{-\infty}^0 xf_X(x)~dx&=\int_\infty^0(-x)f_X(-x)~d(-x)\\ &=\int_0^\infty(-x)f_X(x)~dx\\ &=-\int_0^\infty xf_X(x)~dx\;, \end{align*}$
and therefore $\mu=\int_{-\infty}^0 xf_X(x)~dx+\int_0^\infty xf_X(x)~dx=0\;.$ Thus,
$\begin{align*} \mathrm{Var}(X)&=\int_{-1}^1x^2\left(\frac{p+1}2\right)|x|^p~dx\\ &=\frac{p+1}2\int_{-1}^1 x^2|x|^p~dx\\ &=(p+1)\int_0^1 x^{p+2}~dx\\ &=\frac{p+1}{p+3}\;. \end{align*}$
Now it’s just a first-semester calculus problem in maximizing and minimizing the function $v(p)=\frac{p+1}{p+3}=1-\frac2{p+3}$ over the interval $\left[-\frac12,\frac12\right]$. This is easy: $v'(p)=\frac2{(p+3)^2}>0$ over the entire interval, so the variance is increasing over the entire interval. Thus, it must have its minimum value at $p=-\frac12$ and its maximum at $p=\frac12$. (Actually, you shouldn’t even need any calculus to see that $v(p)$ is increasing.)