To answer the questions succinctly:

$1.$ Number of bits reserved for the Line field in a direct map cache:

The size of the cache is $64$KB$,$ which is $64$ x $1024$ bytes. Each block holds $8$ words, and each word is $4$ bytes. Therefore, each block is $8$ x $4$ bytes $=32$ bytes.

To find the number of lines in the cache, divide the cache size by the block size:

Number of lines $=(64$ x $1024)/32=2048$ lines

To address $2048$ lines, we need log$2(2048)$ bits. So$,$ the number of bits for the Line field is:

log$2(2048)=11$ bits

The correct answer for the Line field is $11.$

$2.$ Number of bits reserved for the Tag field:

A system with $64$GB memory means there are $64$ x $1024^3$ bytes in total. Since each word is $4$ bytes, the total number of words will be $(64$ x $1024^3)/4.$ The total number of addressable blocks in memory is this number divided by the number of words per block $(8).$

Number of blocks $=(64$ x $1024^3)/(8$ x $4)=2^36/2^5=2^31$ blocks

Thus, the main memory address must accommodate $31$ bits for block address. Since we have $11$ bits for the Line field and the word offset $($with $8$ words per block, we need $3$ bits, because $2^3=8),$ we subtract these from the $31$ bits for the total address space:

Tag field bits $=31$ bits $($total$)-11$ bits $($line$)-3$ bits $($word offset$)=17$ bits

The correct answer for the Tag field in a direct map cache is $17($not provided in the options given$).$

$3.$ Main memory block #$4095$ will map to which cache line:

In a direct map cache, to find where a memory block will map, we use the remainder of the division of the block number by the number of lines in the cache.

Cache line number $=$ Block number $\%$ Number of cache lines

Cache line number $=4095\%2048=1023$

The correct answer is $1023.$

$4.$ Number of bits reserved for the Tag field for a $16-$way associative cache:

We still have the same total number of blocks in memory, which is $2^31.$ However, since it's a $16-$way set associative cache, the number of sets will be the total number of lines divided by $16$:

Number of sets $=$ Number of lines $/16=2048/16=128$ sets

To address $128$ sets, we need log$2(128)$ bits, which is $7$ bits. The word offset is still $3$ bits.

So for the Tag field:

Tag field bits $=31$ bits $($total$)-7$ bits $($set$)-3$ bits $($word offset$)=21$ bits

The correct answer for the Tag field in a $16-$way set associative cache is $21($not provided in the options given$).$