a. In order for [tex]f[/tex] to be a PDF, [tex]f[/tex] needs to be non-negative (it is) and the integral over its support must evaluate to 1. [tex]f[/tex] is symmetric about [tex]x=0[/tex], i.e. even, so
[tex]\displaystyle\int_{-\infty}^\infty\frac\lambda2e^{-\lambda|x|}\,\mathrm dx=\lambda\int_0^\infty e^{-\lambda x}\,\mathrm dx=-e^{-\lambda x}\bigg|_0^\infty=1[/tex]
and so [tex]f[/tex] is indeed a PDF.
b. Let [tex]X[/tex] be a random variable with [tex]f[/tex] as its PDF. Then
[tex]\mu=E[X]=\displaystyle\int_{-\infty}^\infty xf(x)\,\mathrm dx=\frac\lambda2\int_{-\infty}^\infty xe^{-\lambda|x|}\,\mathrm dx[/tex]
The integrand is odd, so the integral vanishes and the mean is [tex]\boxed{\mu=0}[/tex].
The variance of [tex]X[/tex] is
[tex]\sigma^2=E[(X-E[X])^2]=E[X^2]-E[X]^2[/tex]
The second moment is
[tex]E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f(x)\,\mathrm dx[/tex]
This integrand is even, so
[tex]E[X^2]=\displaystyle2\int_0^\infty x^2f(x)\,\mathrm dx=\lambda\int_0^\infty x^2e^{-\lambda x}\,\mathrm dx[/tex]
Integrate by parts, taking
[tex]u=x^2\implies\mathrm du=2x\,\mathrm dx[/tex]
[tex]\mathrm dv=e^{-\lambda x}\,\mathrm dx\implies v=-\dfrac{e^{-\lambda x}}\lambda[/tex]
so that
[tex]\displaystyle E[X^2]=\lambda\left(-\frac{x^2e^{-\lambda x}}\lambda\bigg|_0^\infty+2\int_0^\infty\frac{xe^{-\lambda x}}\lambda\,\mathrm dx\right)=2\int_0^\infty xe^{-\lambda x}\,\mathrm dx[/tex]
Integrate by parts again, this time with
[tex]u=x\implies\mathrm du=\mathrm dx[/tex]
[tex]\mathrm dv=e^{-\lambda x}\,\mathrm dx\implies v=-\dfrac{e^{-\lambda x}}\lambda[/tex]
[tex]\displaystyle E[X^2]=2\left(-\frac{xe^{-\lambda x}}\lambda\bigg|_0^\infty+\int_0^\infty\frac{e^{-\lambda x}}\lambda\,\mathrm dx\right)=\frac2\lambda\int_0^\infty e^{-\lambda x}\,\mathrm dx=-\frac2{\lambda^2}e^{-\lambda x}\bigg|_0^\infty=\frac2{\lambda^2}[/tex]
and so the variance is [tex]\boxed{\sigma^2=\frac2{\lambda^2}}[/tex].
