Answer
(a) $f_1 = 156~Hz$
$\lambda_1 = 1.60~m$
(b) F = 467 N
Work Step by Step
(a) Let $f_1$ be the fundamental frequency.
$n~f_1 = 624~Hz$, for some whole number $n$.
$(n+1)~f_1 = 780~Hz$, for some whole number $n$.
We can subtract the first equation from the second equation.
$f_1 = 780~Hz-624~Hz = 156~Hz$
We can find the wavelength $\lambda_1$.
$\lambda_1 = 2L$
$\lambda_1 = (2)(0.800~m)$
$\lambda_1 = 1.60~m$
(b) We can find the speed of the wave along the string.
$v = f_1~\lambda_1$
$v = (156~Hz)(1.60~m)$
$v = 249.6~m/s$
We can find the tension $F$.
$\sqrt{\frac{F}{\mu}} = v$
$F = v^2~\mu$
$F = (249.6~m/s)^2(0.00750~kg/m)$
$F = 467~N$