Sol: $4^{23}$ = $(2^2)^{23}$ = $2^{46}$
=> $4^{23} * 3^{46}$ = $2^{46} * 3^{46}$ = $(2*3)^{46}$ = $6^{46}$
Cyclicity of units' digit of power of 6 is 1
[Go through this post to MASTER Cyclicity of Units' digit of numbers from 2-9 ]

=> Units’ digit of $4^{23} * 3^{46}$ = 6

Sol: Cyclicity of units' digit of power of 2 and 3 is 4
Units' digit of $2^{34}$ = Units' digit of $2^{2}$ (as remainder of 34 by 4 is 2) = 4
Units' digit of $3^{45}$ = Units' digit of $3^{1}$ (as remainder of 45 by 4 is 2) = 1
=> Units’ digit of $2^{34} * 3^{45}$ = 4 * 1 = 4

Sol: Whenever we have to find units digit of power of a big number then we just need to focus on the units' digit of the number and take its power.
=> Units' digit of $1452^{48} * 25463^{123} * 798^{241}$ = Units' digit of $2^{48} * 3^{123} * 8^{241}$
Cyclicity of units' digit of power of 2, 3 and 8 is 4
Units' digit of $2^{48}$ = Units' digit of $2^{4}$ (as remainder of 48 by 4 is 0 so we take unit's digit of the power of the cycle, which is 4) = 6
Units' digit of $3^{123}$ = Units' digit of $3^{3}$ (as remainder of 123 by 4 is 3) = 7
Units' digit of $8^{241}$ = Units' digit of $8^{1}$ (as remainder of 241 by 4 is 1) = 8
=> Units’ digit of $1452^{48} * 25463^{123} * 798^{241}$ = 6 * 7 * 8 = 42 * 8 = ...6

Sol: Units' digit of $145^{2367} * 156^{10987}$ = Units' digit of $5^{2367} * 6^{10987}$
=> Both 5 and 6 have a cycle of 1
=> Units' digit of $145^{2367} * 156^{10987}$ = 5 * 6 = _0

Sol: Units' digit of $224^{12987} * 28569^{1879}$ = Units' digit of $4^{12987} * 9^{1879}$
Cyclicity of units' digit of power of 4 and 9 is 2
=> Units' digit of $4^{12987}$ = Units' digit of $4^1$ = 4
=> Units' digit of $9^{1879}$ = Units' digit of $9^1$ = 9
=> Units’ digit of $224^{12987} * 28569^{1879}$ = 4 * 9 = 6