JavaScript (ES6), 43 bytes

f=(D,t,d)=>5>>d&t>749&t<811&&!f[D]*(f[D]=1)

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Input

• the date as a string in yyyymmdd format


• the time as a string in hhmm format


• the day of week as a 0-indexed integer, with 0 = Tuesday, 1 = Wednesday, ..., 6 = Monday

Output

Returns 0 or 1.

Commented

f = ( // named function, as the underlying object will be used as storage
 D, // D = date (string)
 t, // t = time (string)
 d // d = day of week (integer)
) => //
 5 // 5 is 0000101 in binary, where 1's are set for Tuesday and Thursday
 >> d & // test the relevant bit for the requested day of week
 t > 749 & // test whether we are in the correct time slot
 t < 811 //
 && !f[D] * // make sure that this date was not already invoked at a correct time
 (f[D] = 1) // and store it in the underlying object of f()

APL (Dyalog Unicode), 61 53 50 48 37 36 bytes^SBCS

Anonymous infix lambda. Called with YYYYMMDD f hhmm and then prompts for weekday number; 2 and 4 are Tuesday and Thursday. Redefines the global D to remember dates.

D←⍬
780-⍵

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D←⍬ initialise D to be an empty set

… anonymous lambda; ⍺ is YYYYMMDD, ⍵ is hhmm

 780-⍵ difference between 780 (mean of 0750 and 0810) and the time

 | absolute value of that

 30≥ is 30 greater or equal to that?

 (…)< and it is not true that:

  ⍺∊D the date is a member of D

 (…)∧ and it is true that:

  ⎕∊2 4 the prompted for weekday is a member of the set 2,4

 ⍺/⍨ use that to compress the date (i.e. gives if false, date if true)

 D,← append that to D

 ≢ and return its tally (i.e. 0 or 1, which are APL's false and true)

Python 3, 69 bytes

f=lambda w,r,*t,l=0:r not in l!=w in(2,4)<(7,49)<t<(8,11)!=l.add(r)

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Takes input as f(day of the week, date, hours, minutes), where date can be in any consistent format.

R, 114 106 bytes

function(D,y=D:"%D",u=D$h==7&D$mi>49|D$h==8&D$mi<11&D$w%in%2^4&!y%in%L,`:`=format,`^`=c)L<<-L^y[u];u
L=F

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Persistence:

Date is checked against L, the list of dates where the code returned TRUE. When the code returns TRUE, today's date is appended to this list. Otherwise the list is not modified.

Saved 6 bytes thanks to @Giuseppe!

Made the code actually work and saved 2 bytes thanks to @digEmAll!

Excel formula, 85 bytes

=IF(AND(MID(WEEKDAY(A1)/0,684;3;1)="3";A1-INT(A1)>=0,32638;A1-INT(A1)<=0,34028);TRUE)

Weekday with no parameters are from 1 (Sunday) to 7 (Saturday). The days we want are 3 and 5. Dividing all numbers from 1 to 7 for 0,648, only 3 and 5 gives a result where the first decimal is 3 (Got it by dividing with rand())

Input is inserted on Cell A1

Clean, 343 326 303 279 216 bytes

Clean is so ill-suited to this it's like trying to paint a fence with a chainsaw.

import StdEnv,System.Environment,System._Unsafe
?(y,x,z)=y*480+x*40+z
$y h m d=(d-3)^2==1&&((h-8)*60+m)^2<121&&appUnsafe(setEnvironmentVariable"l"(fromInt(?y)))(maybe 0toInt(accUnsafe(getEnvironmentVariable"l")))< ?y

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Golfing then explaination.

C (gcc),  78   50  49 bytes

D;f(d,w,t)w=d-D&&w<4&&w%2&&t>749&&t<811&&(D=d);

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The expected inputs are:

• 
  d: the date, as a single number yyyymmdd
  


• 
  w: the day of the week, starting with Monday (0)


• 
  t: the time, as a single number hhmm
  

Explanation

D; // the date we last said hello.
f(d, // date
 w, // day of the week
 t) // time

 w= // replaces return
 d-D // if we did not say hello today
 &&w<4&&w%2 // and we are Tuesday(1) or Thursday(3)
 &&t>749&&t<811 // and time is between 7:50 and 8:10, inclusive
 &&(D=d); // then we say hello (evaluated to true) and update D

Edits

• Saved 28 bytes thanks to Adám


• Saved 1 more byte, since abs() was actually not helping with the new version

Batch, 109 bytes

@if %3 neq 2 if %3 neq 4 exit/b1
@if %2 geq 07:50 if %2 leq 08:10 if .%1 neq .%l% set l=%1&exit/b0
@exit/b1

Takes input in the form date time dow e.g. 2018-09-04 08:10 2 and outputs via exit code. Explanation: The environment variable l (or any other single letter would work) is used to store the last successful date that passes the test. (The date format itself does not matter as long as it is consistent and does not contain spaces.)

Perl 6, 33 bytes

811>$^t>749>5+>$^w%2>(%)$^d++

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Heavily inspired by Arnauld's solution. Uses the same input format.

C#, 121 Bytes

int d=new int2,4;
double s=>Now.TimeOfDay.TotalSeconds;
bool h=>d.Contains((int)Now.DayOfWeek)&&s>=470&&s<=490;

Moving all three to the same line reduces size to 117 bytes. h is used as a property, just read the value prior to sending the message:

if (h) SendMessage();

F#, 119 bytes

let f w d h m l = if not(l|>Seq.contains d)&&[3;5]|>Seq.contains w&&(h=7&&m>49||h=8&&m<11)then(l@[d],true)else(l,false)

let f w d h m l =

declare function called f with parameters w (day of week) d (date) h (hour) m (minute) l (list of dates it's run on)

if not(l|>Seq.contains d)
if the list of dates doesn't contain the passed date

&&[3;5]|>Seq.contains w and the day is Tuesday (3) or Wednesday (5)

&&(h=7&&m>49||h=8&&m<11) and the time is between (exclusive) 7:49 and 8:11

then(l@[d],true) then return a tuple containing the list of dates with the current date appended, and true

else(l,false) else return a tuple containing the list of dates without today and false

Bash 95 87 bytes

[ $(ps -ef|grep $0|wc -l) -gt 3 -o $2 -lt 750 -o $2 -gt 810 ]||((10>>$3&1))&&sleep 20m

Edit: saved 8 bytes by stealing an idea from Annyo

Call with yyyymmdd HHMM D, with D starting at 0, for Monday.

I'm not entirely sure why I have the line count check set to three, it seems it should be two, but two fails to work and three seems to work.

Also, I haven't bothered using the date, but the requirements seem to indicate that we should accept the date, so I took the option mentioned to take the time separately and just ignore the date value.