Here, mantissa is represented in normalized representation and exponent in $\text{excess-}64$ (subtract $64$ to get actual value).
a. We have to represent $-(7.5)_{10} = -(111.1)_2$.
Now we are using base $16$ for exponent. So, mantissa will be $.01111$ and this makes exponent as $1 (4$ bit positions and no hiding first $1$ as in $\text{IEEE}\; 754$ as this is not mentioned in question) which in $\text{excess-}64$ will be $64 + 1 = 65.$ Number being negative sign bit is $1.$ So, we get
$(1\; 01111\underbrace{000\dots0}_{\text{19 zeroes}} \; 1000001)_2 = (\text{BC}000041)_{16}$
b. Largest value will be with largest possible mantissa, largest possible exponent and positive sign bit. So, this will be all $1$'s except sign bit which will be
$0.\underbrace{111\dots 1}_{\text{24 ones}} \times 16^{127-64} = \left(1- 2^{-24}\right) \times 16^{63}$
$($Again we did not add implicit $1$ as in $\text{IEEE}\; 754)$
$2^x = 10^{y} \implies y = \log 2^x = x \log 2$
So, $(1-2^{-24}) \times 16^{63} \\= \left(1 - 10^{-24 \log 2}\right) \times 10^{63 \log 16}\\ \approx \left(1 - 10^{-7}\right) \times 10^{76}\\ = 10^{76}$
Not directly relevant here, but a useful read: https://jeapostrophe.github.io/courses/2015/fall/305/notes/dist/reading/help-floating-point.pdf