Given:
ΔT = 38 - 26 = 12°C, temperature change
Q = 11.3 J, heat input
c = 0.128 J/(g-°C), specific heat of lead
Let m = the mass of the lead.
Then
Q = m*c*ΔT
(m g)*(0.128 J/(g-°C))*(12 °C) = 11.3 J
1.536m = 11.3
m = 7.357 g
Answer: 7.36 g (2 sig. figs)
The two different isotopes have weights :
w1 = 78.918 amu
w2 = 80.916 amu
average weight w3 = 79.903 amu
The mixing of two components can be modeled as
let the fraction of w1 be 'x'
hence 
now this is a linear equation in 'x'. Substituting the values we get
x = 0.507
hence the percentage of Br79 = 50.7% and the percentage of BR81 = 49.3%