I submit the following calculations in the hope that it will help you.
µ₁=785/18=43.62, µ₂=770/13=59.23.
s₁=40, s₂=25, n₁=18, n₂=13.
s₁²/n₁=1600/18=88.89, s₂²/n₂=625/13=48.07.
SE (standard error) for joint sample=√(s₁²/n₁+s₂²/n₂)=11.70.
DOF (degrees of freedom)=SE²/((s₁²/n₁)/(n₁-1)+(s₂²/n₂)/(n₂-1))=15 to the nearest integer. Because we are testing for a one-sided difference (µ₁>µ₂ rather than µ₁≠µ₂, or µ₁<>µ₂) we need to apply a 1-tail test at 0.01 significance level. t value from tables is 2.602 for 15 dof.
Now we need to calculate a t value to compare with.
t(calc)=(µ₁-µ₂)/SE=(43.62-59.23)/11.70=-1.33. The sign is irrelevant, so we compare 1.33 with 2.60. The calculated t value is less than the critical t value therefore we can’t reject the null hypothesis that µ₁>µ₂.